Learn4Less Logo

Navigation

Integration

1.3 The Fundamental Theorem of Calculus

52 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1Past exam · 2016Q2

Suppose that f(x)f(x) is a function and F(x)=e(x2−3)+1F(x) = e^{(x^2-3)} + 1 is an antiderivative of f(x)f(x). Evaluate the definite integral ∫15f(x) dx\displaystyle\int_1^{\sqrt5} f(x)\,\dee{x}.

Answer

e2−e−2e^2-e^{-2}

Full solution

The Fundamental Theorem of Calculus Part 2 (Theorem 1.3.1 in the CLP-2 text) tells us that

∫15f(x) dx=F(5)−F(1)=(e(52−3)+1)−(e(12−3)+1)=e5−3−e1−3=e2−e−2\begin{align*} \int_1^{\sqrt5} f(x)\,\dee{x} &= F(\sqrt5) - F(1) \\ &= \big( e^{(\sqrt5^2-3)} + 1 \big) - \big( e^{(1^2-3)} + 1 \big) \\ &= e^{5-3} - e^{1-3} = e^2-e^{-2} \end{align*}
Q2Stage 1Past exam · M105 2015A

For the function f(x)=x3−sin⁡2xf(x) = x^3 -\sin 2x, find its antiderivative F(x)F(x) that satisfies F(0)=1F(0)=1.

Hint

First find the general antiderivative by guessing and checking.

Answer

F(x)=x44+12cos⁡2x+12F(x) = \dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+\dfrac{1}{2}.

Full solution

First, let's find a general antiderivative of x3−sin⁡(2x)x^3-\sin(2x).

  • One function with derivative x3x^3 is x44\dfrac{x^4}{4}.

  • To find an antiderivative of sin⁡(2x)\sin(2x), we might first guess cos⁡(2x)\cos(2x); checking, we see ddx{cos⁡(2x)}=−2sin⁡(2x)\diff{}{x}\{\cos(2x)\}=-2\sin(2x). So, we only need to multiply by −12-\dfrac{1}{2}: ddx{−12cos⁡2x}=sin⁡(2x)\displaystyle\diff{}{x}\left\{-\dfrac{1}{2}\cos 2x\right\}=\sin(2x).

So, the general antiderivative of f(x)f(x) is x44+12cos⁡2x+C\dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+C. To satisfy F(0)=1F(0)=1, we need (The symbol   ⟺  \iff is read “if and only if”. This is used in mathematics to express the logical equivalence of two statements. To be more precise, the statement P  ⟺  QP \iff Q tells us that PP is true whenever QQ is true and QQ is true whenever PP is true.)

[x44+12cos⁡2x+C]x=0=1  ⟺  12+C=1  ⟺  C=12\begin{align*} \Big[\frac{x^4}{4}+\frac{1}{2}\cos 2x+C\Big]_{x=0}=1 \iff \frac{1}{2} + C = 1 \iff C=\frac{1}{2} \end{align*}

So F(x)=x44+12cos⁡2x+12F(x) = \dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+\dfrac{1}{2}.

Q3Stage 1Past exam · 2014D

Decide whether each of the following statements is true or false. Provide a brief justification.

  1. If f(x)f(x) is continuous on [1,π][1, \pi] and differentiable on (1,π)(1,\pi), then ∫1πf′(x) dx=f(π)−f(1)\displaystyle\int_1^\pi f'(x)\,\dee{x} = f(\pi)-f(1).

  2. ∫−111x2 dx=0\displaystyle\int_{-1}^1 \frac{1}{x^2}\,\dee{x} = 0.

  3. If ff is continuous on [a,b][a, b] then ∫abxf(x) dx=x∫abf(x) dx\displaystyle\int_a^b xf(x)\,\dee{x} = x\int_a^b f(x)\,\dee{x}.

Hint

Be careful. Two of these make no sense at all.

Answer

(a) True (b) False (c) False, unless ∫abf(x) dx=∫abxf(x) dx=0\int_a^b f(x)\,\dee{x}=\int_a^b xf(x)\,\dee{x} = 0.

Full solution

(a) This is true, by part 2 of the Fundamental Theorem of Calculus, Thereom 1.3.1 in the CLP-2 text with G(x)=f(x)G(x)=f(x) and f(x)f(x) replaced by f′(x)f'(x).

(b) This is not only false, but it makes no sense at all. The integrand is strictly positive so the integral has to be strictly positive. In fact it's +∞+\infty. The Fundamental Theorem of Calculus does not apply because the integrand has an infinite discontinuity at x=0x=0.

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

(c) This is not only false, but it makes no sense at all, unless ∫abf(x) dx=∫abxf(x) dx=0\int_a^b f(x)\,\dee{x}=\int_a^b xf(x)\,\dee{x} = 0. The left hand side is a number. The right hand side is a number times xx.

∫abxf(x) dx⏟areavsx⏟variable⋅∫abf(x) dx⏟area\underbrace{\int_a^b xf(x)\ \dee{x}}_{\mathrm{area}} \qquad\text{vs}\qquad \underbrace{x}_{\mathrm{variable}}\cdot\underbrace{\int_a^bf(x)\ \dee{x}}_{\mathrm{area}}

For example, if a=0a=0, b=1b=1 and f(x)=1f(x) = 1, then the left hand side is ∫01x dx=12\int_0^1 x\,\dee{x} = \frac{1}{2} and the right hand side is x∫01dx=xx\int_0^1 \dee{x}=x.

Q4Stage 1

True or false: an antiderivative of 1x2\dfrac{1}{x^2} is log⁡(x2)\log (x^2) (where by log⁡x\log x we mean logarithm base ee).

Hint

Check by differentiating.

Answer

false

Full solution

This is a tempting thought:

∫1x dx=log⁡∣x∣+C\begin{align*}\int \frac{1}{x}\ \dee{x}&=\log|x|+C\end{align*}

so perhaps similarly

∫1x2 dx=?log⁡∣x2∣+C=log⁡(x2)+C\begin{align*}\int \frac{1}{x^2}\ \dee{x}&\stackrel{?}{=}\log|x^2|+C=\log(x^2)+C\end{align*}

We check by differentiating:

ddx{log⁡(x2)}=ddx{2log⁡x}=2x≠1x2\begin{align*}\diff{}{x}\{\log(x^2)\} &= \diff{}{x}\{2\log x\}=\frac{2}{x} \neq \frac{1}{x^2}\end{align*}

So, it wasn't so easy: false.

When we're guessing antiderivatives, we often need to adjust our original guesses a little. Changing constants works well; changing functions usually does not.

Q5Stage 1

True or false: an antiderivative of cos⁡(ex)\cos(e^x) is sin⁡(ex)ex\frac{\sin(e^x)}{e^x}.

Hint

Check by differentiating.

Answer

false

Full solution

This is tempting:

ddx{sin⁡(ex)}=excos⁡(ex)\begin{align*}\diff{}{x}\{\sin(e^x)\} &= e^x\cos(e^x)\end{align*}

so perhaps

ddx{sin⁡(ex)ex}=?cos⁡(ex)\begin{align*}\diff{}{x}\left\{\frac{\sin(e^x)}{e^x}\right\} &\stackrel{?}{=} \cos(e^x)\end{align*}

We check by differentiating:

ddx{sin⁡(ex)ex}=ex(cos⁡(ex)⋅ex)−sin⁡(ex)exe2x(quotient rule)=cos⁡(ex)−sin⁡(ex)ex≠cos⁡(ex)\begin{align*}\diff{}{x}\left\{\frac{\sin(e^x)}{e^x}\right\} &=\frac{e^x\left(\cos(e^x)\cdot e^x\right)-\sin(e^x)e^x}{e^{2x}} &\text{(quotient rule)}\\ & = \cos (e^x) - \frac{\sin(e^x)}{e^x}\\ &\neq \cos(e^x)\end{align*}

So, the statement is false.

When we're guessing antiderivatives, we often need to adjust our original guesses a little. Dividing by constants works well; dividing by functions usually does not.

Q6Stage 1

Suppose F(x)=∫7xsin⁡(t2) dtF(x) = \displaystyle\int_7^x \sin(t^2)\ \dee{t}. What is the instantaneous rate of change of F(x)F(x) with respect to xx?

Hint

Use the Fundamental Theorem of Calculus Part 1.

Answer

sin⁡(x2)\sin(x^2)

Full solution

“The instantaneous rate of change of F(x)F(x) with respect to xx" is another way of saying “F′(x)F'(x)". From the Fundamental Theorem of Calculus Part 1, we know this is sin⁡(x2)\sin (x^2).

Q7Stage 1

Suppose F(x)=∫2xe1/t dtF(x) = \displaystyle\int_{2}^x e^{1/t}\ \dee{t}. What is the slope of the tangent line to y=F(x)y=F(x) when x=3x=3?

Hint

Use the Fundamental Theorem of Calculus, Part 1.

Answer

e3\sqrt[3]{e}

Full solution

The slope of the tangent line to y=F(x)y=F(x) when x=3x=3 is exactly F′(3)F'(3). By the Fundamental Theorem of Calculus Part 1, F′(x)=e1/xF'(x) = e^{1/x}. Then F′(3)=e1/3=e3F'(3) = e^{1/3} = \sqrt[3]{e}.

Q8Stage 1

Suppose F′(x)=f(x)F'(x)=f(x). Give two different antiderivatives of f(x)f(x).

Hint

You already know that F(x)F(x) is an antiderivative of f(x)f(x).

Answer

For any constant CC, F(x)+CF(x)+C is an antiderivative of f(x)f(x). So, for example, F(x)F(x) and F(x)+1F(x)+1 are both antiderivatives of f(x)f(x).

Full solution

For any constant CC, F(x)+CF(x)+C is an antiderivative of f(x)f(x), because ddx{F(x)+C}=ddx{F(x)}=f(x)\diff{}{x}\{F(x)+C\} = \diff{}{x}\{F(x)\} = f(x). So, for example, F(x)F(x) and F(x)+1F(x)+1 are both antiderivatives of f(x)f(x).

Q9Stage 1

In Question 45, Section 1.1, we found that

∫0a1−x2 dx=π4−12arccos⁡(a)+12a1−a2.\int_0^a\sqrt{1-x^2}\ \dee{x}=\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}.

  1. Verify that dda{π4−12arccos⁡(a)+12a1−a2}=1−a2\displaystyle\diff{}{a}\left\{\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}\right\} = \sqrt{1-a^2}.

  2. Find a function F(x)F(x) that satisfies F′(x)=1−x2F'(x) = \sqrt{1-x^2} and F(0)=πF(0)=\pi.

Hint

(a) Recall ddx{arccos⁡x}=−11−x2\diff{}{x}\{\arccos x\} = \frac{-1}{\sqrt{1-x^2}}.
(b) All antiderivatives of 1−x2\sqrt{1-x^2} differ from one another by a constant. You already know one antiderivative.

Answer

  1. We differentiate with respect to aa. Recall ddx{arccos⁡x}=−11−x2\diff{}{x}\{\arccos x\} = \frac{-1}{\sqrt{1-x^2}}. To differentiate 12a1−a2\frac{1}{2}a\sqrt{1-a^2}, we use the product and chain rules.

    dda{π4−12arccos⁡(a)+12a1−a2}=0−12⋅−11−a2+(12a)⋅−2a21−a2+121−a2=121−a2−a221−a2+1−a221−a2=1−a2+1−a221−a2=2(1−a2)21−a2=1−a2\begin{align*} \diff{}{a}\left\{\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}\right\} &= 0-\frac{1}{2}\cdot\frac{-1}{\sqrt{1-a^2}} + \left(\frac{1}{2}a\right)\cdot\frac{-2a}{2\sqrt{1-a^2}} + \frac{1}{2}\sqrt{1-a^2}\\ &=\frac{1}{2\sqrt{1-a^2}}- \frac{a^2}{2\sqrt{1-a^2}}+\frac{1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{1-a^2+1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{2(1-a^2)}{2\sqrt{1-a^2}} \\&=\sqrt{1-a^2} \end{align*}

  2. F(x)=5π4−12arccos⁡(x)+12x1−x2F(x) = \dfrac{5\pi}{4}-\dfrac{1}{2}\arccos(x)+\dfrac{1}{2}x\sqrt{1-x^2}

Full solution

  1. We differentiate with respect to aa. Recall ddx{arccos⁡x}=−11−x2\diff{}{x}\{\arccos x\} = \frac{-1}{\sqrt{1-x^2}}. To differentiate 12a1−a2\frac{1}{2}a\sqrt{1-a^2}, we use the product and chain rules.

    dda{π4−12arccos⁡(a)+12a1−a2}=0−12⋅−11−a2+(12a)⋅−2a21−a2+121−a2=121−a2−a221−a2+1−a221−a2=1−a2+1−a221−a2=2(1−a2)21−a2=1−a2\begin{align*} \diff{}{a}\left\{\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}\right\} &= 0-\frac{1}{2}\cdot\frac{-1}{\sqrt{1-a^2}} + \left(\frac{1}{2}a\right)\cdot\frac{-2a}{2\sqrt{1-a^2}} + \frac{1}{2}\sqrt{1-a^2}\\ &=\frac{1}{2\sqrt{1-a^2}}- \frac{a^2}{2\sqrt{1-a^2}}+\frac{1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{1-a^2+1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{2(1-a^2)}{2\sqrt{1-a^2}} \\&=\sqrt{1-a^2} \end{align*}

  2. Let G(x)=π4−12arccos⁡(x)+12x1−x2G(x) = \frac{\pi}{4}-\frac{1}{2}\arccos(x)+\frac{1}{2}x\sqrt{1-x^2}. We showed in part (a) that G(x)G(x) is an antiderivative of 1−x2\sqrt{1-x^2}. Since F(x)F(x) is also an antiderivative of 1−x2\sqrt{1-x^2}, F(x)=G(x)+CF(x) = G(x)+C for some constant CC (this is Lemma 1.3.8 in the CLP-2 text).

    Note G(0)=∫001−x2 dx=0G(0)=\displaystyle\int_0^0\sqrt{1-x^2}\ \dee{x} =0, so if F(0)=πF(0)=\pi, then F(x)=G(x)+πF(x)=G(x)+\pi. That is,

    F(x)=5π4−12arccos⁡(x)+12x1−x2 .F(x) = \frac{5\pi}{4}-\frac{1}{2}\arccos(x)+\frac{1}{2}x\sqrt{1-x^2}\ .
Q10Stage 1

Evaluate the following integrals using the Fundamental Theorem of Calculus Part 2, or explain why it does not apply.

  1. ∫−ππcos⁡x dx\displaystyle\int_{-\pi}^\pi \cos x \ \dee{x}.

  2. ∫−ππsec⁡2x dx\displaystyle\int_{-\pi}^\pi \sec^2 x \ \dee{x}.

  3. ∫−201x+1 dx\displaystyle\int_{-2}^0 \frac{1}{x+1}\ \dee{x}.

Hint

In order to apply the Fundamental Theorem of Calculus Part 2, the integrand must be continuous over the interval of integration.

Answer

(a) 0 (b),(c) The FTC does not apply, because the integrand is not continuous over the interval of integration.

Full solution

  1. The antiderivative of cos⁡x\cos x is sin⁡x\sin x, and cos⁡x\cos x is continuous everywhere, so ∫−ππcos⁡x dx=sin⁡(π)−sin⁡(−π)=0\displaystyle\int_{-\pi}^\pi \cos x \ \dee{x} = \sin(\pi)-\sin(-\pi) = 0.

  2. Since sec⁡2x\sec^2 x is discontinuous at x=±π2x=\pm\frac{\pi}{2}, the Fundamental Theorem of Calculus Part 2 does not apply to ∫−ππsec⁡2x dx\displaystyle\int_{-\pi}^\pi \sec^2 x \ \dee{x}.

  3. Since 1x+1\frac{1}{x+1} is discontinuous at x=−1x=-1, the Fundamental Theorem of Calculus Part 2 does not apply to ∫−201x+1 dx\displaystyle\int_{-2}^0 \frac{1}{x+1}\ \dee{x}.

Questions 11 through 14 are meant to help reinforce key ideas in the Fundamental Theorem of Calculus and its proof.

Q11Stage 1

As in the proof of the Fundamental Theorem of Calculus, let F(x)=∫axf(t) dtF(x) = \int_{a}^x f(t)\ \dee{t}. In the diagram below, shade the area corresponding to F(x+h)−F(x)F(x+h)-F(x).

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Hint

Use the definition of F(x)F(x) as an area.

Answer

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Full solution

Using the definition of FF, F(x)\textcolor{red}{F(x)} is the area under the curve from aa to xx, and F(x+h)\textcolor{blue}{F(x+h)} is the area under the curve from aa to x+hx+h. These are shown on the same diagram, below.

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Then the area represented by F(x+h)−F(x)\textcolor{blue}{F(x+h)}-\textcolor{red}{F(x)} is the area that is outside the red, but inside the blue. Equivalently, it is ∫xx+hf(t) dt\int\limits_{x}^{x+h} f(t)\ \dee{t}.

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Q12Stage 1

Let F(x)=∫0xf(t)dtF(x) = \displaystyle\int_0^x f(t)\dee{t}, where f(t)f(t) is shown in the graph below, and 0≤x≤40 \leq x \leq 4.

  1. Is F(0)F(0) positive, negative, or zero?

  2. Where is F(x)F(x) increasing and where is it decreasing?

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Hint

F(x)F(x) represents net signed area.

Answer

(a) zero (b) increasing when 0<x<10 < x < 1 and 3<x<43<x<4; decreasing when 1<x<31<x<3

Full solution

We evaluate F(0)F(0) using the definition: F(0)=∫00f(t) dt=0F(0) = \int_0^0 f(t)\ \dee{t}=0. Although f(0)>0f(0)>0, the area from t=0t=0 to t=0t=0 is zero.
As xx moves along, F(x)F(x) adds bits of signed area. If it's adding positive area, it's increasing, and if it's adding negative area, it's decreasing. So, F(x)F(x) is increasing when 0<x<10 < x < 1 and 3<x<43<x<4, and F(x)F(x) is decreasing when 1<x<31<x<3.

Q13Stage 1

Let G(x)=∫x0f(t)dtG(x) = \displaystyle\int_x^0 f(t)\dee{t}, where f(t)f(t) is shown in the graph below, and 0≤x≤40 \leq x \leq 4.

  1. Is G(0)G(0) positive, negative, or zero?

  2. Where is G(x)G(x) increasing and where is it decreasing?

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Hint

Note G(x)=−F(x)G(x)=-F(x), when F(x)F(x) is defined as in Question 12.

Answer

(a) zero (b) G(x)G(x) is increasing when 1<x<31<x<3, and it is decreasing when 0<x<10<x<1 and when 3<x<43<x<4.

Full solution

This question is nearly identical to Question 12, with

G(x)=∫x0f(t) dt=−∫0xf(t) dt=−F(x).G(x) = \displaystyle\int_x^0 f(t)\ \dee{t} = -\displaystyle\int_0^x f(t)\ \dee{t}=-F(x).

So, G(x)G(x) increases when F(x)F(x) decreases, and vice-versa. Therefore: G(0)=0G(0)=0, G(x)G(x) is increasing when 1<x<31<x<3, and G(x)G(x) is decreasing when 0<x<10<x<1 and when 3<x<43<x<4.

Q14Stage 1

Let F(x)=∫axt dtF(x) = \displaystyle\int_a^x t\ \dee{t}. Using the definition of the derivative, find F′(x)F'(x).

Hint

Using the definition of the derivative, F′(x)=lim⁡h→0F(x+h)−F(x)hF'(x) = \displaystyle\lim_{h \to 0}\dfrac{F(x+h)-F(x)}{h}.

The area of a trapezoid with base bb and heights h1h_1 and h2h_2 is 12b(h1+h2)\frac{1}{2}b(h_1+h_2).

Answer

Using the definition of the derivative,

F′(x)=lim⁡h→0F(x+h)−F(x)h=lim⁡h→0∫ax+ht dt−∫axt dth=lim⁡h→0∫xx+ht dth\begin{align*}F'(x) & = \displaystyle\lim_{h \to 0}\dfrac{F(x+h)-F(x)}{h}\\ &=\lim_{h \to 0}\dfrac{\int_a^{x+h} t\ \dee{t}-\int_a^x t\ \dee{t}}{h}\\ &=\lim_{h \to 0}\dfrac{\int_x^{x+h} t\ \dee{t}}{h}\end{align*}

The numerator describes the area of a trapezoid with base hh and heights xx and x+hx+h.

=lim⁡h→012h(x+x+h)h=lim⁡h→0(x+12h)=x\begin{align*}&=\lim_{h \to 0}\dfrac{\frac{1}{2}h(x+x+h)}{h}\\ &=\lim_{h \to 0}\left(x+\frac{1}{2}h\right)\\ &=x\end{align*}

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

So, F′(x)=xF'(x)=x.

Full solution

Using the definition of the derivative,

F′(x)=lim⁡h→0F(x+h)−F(x)h=lim⁡h→0∫ax+ht dt−∫axt dth=lim⁡h→0∫xx+ht dth\begin{align*}F'(x) & = \displaystyle\lim_{h \to 0}\dfrac{F(x+h)-F(x)}{h}\\ &=\lim_{h \to 0}\dfrac{\int_a^{x+h} t\ \dee{t}-\int_a^x t\ \dee{t}}{h}\\ &=\lim_{h \to 0}\dfrac{\int_x^{x+h} t\ \dee{t}}{h}\end{align*}

The numerator describes the area of a trapezoid with base hh and heights xx and x+hx+h.

=lim⁡h→012h(x+x+h)h=lim⁡h→0(x+12h)=x\begin{align*}&=\lim_{h \to 0}\dfrac{\frac{1}{2}h(x+x+h)}{h}\\ &=\lim_{h \to 0}\left(x+\frac{1}{2}h\right)\\ &=x\end{align*}

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

So, F′(x)=xF'(x)=x.

Q15Stage 1

Give a continuous function f(x)f(x) so that F(x)=∫0xf(t)dtF(x) = \displaystyle\int_0^x f(t)\dee{t} is a constant.

Hint

There is only one!

Answer

f(t)=0f(t)=0

Full solution

If F(x)F(x) is constant, then F′(x)=0F'(x)=0. By the Fundamental Theorem of Calculus Part 1, F′(x)=f(x)F'(x)=f(x). So, the only possible continuous function fitting the question is f(x)=0f(x)=0.

This makes intuitive sense: if moving xx doesn't add or subtract area under the curve, then there must not be any area under the curve–the curve should be the same as the xx-axis.

As an aside, we mention that there are other, non-continuous functions f(t)f(t) such that ∫0xf(t) dt=0\int_0^x f(t)\ \dee{t} = 0 for all xx. For example, f(t)={0x≠01x=0f(t) = \left\{\begin{array}{cc} 0 & x \neq 0\\ 1 & x=0 \end{array}\right.. These kinds of removable discontinuities will not factor heavily in our discussion of integrals.

So far, we have been able to guess many antiderivatives. Often, however, antiderivatives are very difficult to guess. In Questions 16 through 19, we will find some antiderivatives that might appear in a table of integrals. Coming up with the antiderivative might be quite difficult (strategies to do just that will form a large part of this semester), but verifying that your antiderivative is correct is as simple as differentiating.

Q16Stage 1

Evaluate and simplify ddx{xlog⁡(ax)−x}\diff{}{x}\{x\log(ax)-x\}, where aa is some constant and log⁡(x)\log(x) is the logarithm base ee. What antiderivative does this tell you?

Hint

If ddx{F(x)}=f(x)\diff{}{x}\{F(x)\}=f(x), that tells us ∫f(x) dx=F(x)+C\int f(x)\ \dee{x} = F(x)+C.

Answer

∫log⁡(ax) dx=xlog⁡(ax)−x+C\int \log(ax)\ \dee{x}= x\log(ax)-x+C, where aa is a given constant, and CC is any constant.

Full solution
ddx{xlog⁡(ax)−x}=x(aax)+log⁡(ax)−1(product rule, chain rule)=log⁡(ax)\begin{align*}\diff{}{x}\{x\log(ax)-x\}&=x\left(\frac{a}{ax}\right)+\log(ax)-1 &\text{(product rule, chain rule)}\\ &=\log(ax)\end{align*}

So, we know

∫log⁡(ax) dx=xlog⁡(ax)−x+Cwhere a is a given constant, and C is any constant.\begin{align*}\int \log(ax)\ \dee{x}&= x\log(ax)-x+C&\text{where }a\text{ is a given constant, and }C\text{ is any constant.}\end{align*}

Remark: ∫log⁡(ax) dx\int \log(ax)\ \dee{x} can be calculated using the method of Integration by Parts, which you will learn in Section 1.7 of the CLP-2 text.

Q17Stage 1

Evaluate and simplify ddx{ex(x3−3x2+6x−6)}\diff{}{x}\{e^x\left(x^3-3x^2+6x-6\right)\}. What antiderivative does this tell you?

Hint

When you're differentiating, you can leave the exe^x factored out.

Answer

∫x3ex dx=ex(x3−3x2+6x−6)+C\int x^3e^x\ \dee{x}=e^x\left(x^3-3x^2+6x-6\right)+C

Full solution
ddx{ex(x3−3x2+6x−6)}=ex(3x2−6x+6)+ex(x3−3x2+6x−6)(product rule)=ex(3x2−6x+6+x3−3x2+6x−6)=x3ex\begin{align*}\diff{}{x}\left\{e^x\left(x^3-3x^2+6x-6\right)\right\}&=e^x\left(3x^2-6x+6\right)+e^x\left(x^3-3x^2+6x-6\right) &\text{(product rule)}\\ &=e^x\left(3x^2-6x+6+x^3-3x^2+6x-6\right)\\ &=x^3e^x\end{align*}

So,

∫x3ex dx=ex(x3−3x2+6x−6)+C\begin{align*}\int x^3e^x\ \dee{x}&=e^x\left(x^3-3x^2+6x-6\right)+C\end{align*}

Remark: ∫x3ex dx\int x^3e^x\ \dee{x} can be calculated using the method of Integration by Parts, which you will learn in Section 1.7 of the CLP-2 text.

Q18Stage 1

Evaluate and simplify ddx{log⁡∣x+x2+a2∣}\diff{}{x}\left\{\log\left|x+\sqrt{x^2+a^2}\right|\right\}, where aa is some constant. What antiderivative does this tell you?

Hint

After differentiation, you can simplify pretty far. Keep at it!

Answer

∫1x2+a2 dx=log⁡∣x+x2+a2∣+C\displaystyle\int \dfrac{1}{\sqrt{x^2+a^2}}\ \dee{x} = \log\left|x+\sqrt{x^2+a^2}\right|+C when aa is a given constant. As usual, CC is an arbitrary constant.

Full solution
ddx{log⁡∣x+x2+a2∣}=1x+x2+a2⋅(1+12x2+a2⋅2x)(chain rule)=1+xx2+a2x+x2+a2=x2+a2+xx2+a2x+x2+a2=1x2+a2\begin{align*}\diff{}{x}\left\{\log\left|x+\sqrt{x^2+a^2}\right|\right\}&= \frac{1}{x+\sqrt{x^2+a^2}}\cdot \left(1+\frac{1}{2\sqrt{x^2+a^2}}\cdot 2x\right) &\text{(chain rule)}\\ &=\frac{1+\frac{x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}} =\frac{\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}}\\ &=\frac{1}{\sqrt{x^2+a^2}}\end{align*}

So,

∫1x2+a2 dx=log⁡∣x+x2+a2∣+C\begin{align*}\int \frac{1}{\sqrt{x^2+a^2}}\ \dee{x} &= \log\left|x+\sqrt{x^2+a^2}\right|+C\end{align*}

Remark: ∫1x2+a2 dx\int \frac{1}{\sqrt{x^2+a^2}}\ \dee{x} can be calculated using the method of Trigonometric Substitution, which you will learn in Section 1.9 of the CLP-2 text.

Q19Stage 1

Evaluate and simplify ddx{x(a+x)−alog⁡(x+a+x)}\displaystyle\diff{}{x}\left\{\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)\right\}, where aa is some constant. What antiderivative does this tell you?

Hint

This derivative also simplifies considerably. You might need to add fractions by finding a common denominator.

Answer

∫xx(a+x) dx=x(a+x)−alog⁡(x+a+x)+C\displaystyle\int \dfrac{x}{\sqrt{x(a+x)}}\ \dee{x}=\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)+C

Full solution

Using the chain rule:

ddx{x(a+x)−alog⁡(x+a+x)}=x+(a+x)2x(a+x)−a(1x+a+x⋅(12x+12a+x))=2x+a2x(a+x)−a(1x+a+x⋅(a+x+x2x(a+x)))=2x+a2x(a+x)−a(12x(a+x))=2x2x(a+x)=xx(a+x)\begin{align*}\diff{}{x}&\left\{\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)\right\}\\ &= \frac{x+(a+x)}{2\sqrt{x(a+x)}}-a\left(\frac{1}{\sqrt{x}+\sqrt{a+x}}\cdot\left(\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{a+x}}\right)\right) \\&= \frac{2x+a}{2\sqrt{x(a+x)}}- a\left( \frac{1}{\sqrt{x}+\sqrt{a+x}}\cdot\left( \frac{\sqrt{a+x}+\sqrt{x}}{2\sqrt{x(a+x)}}\right)\right) \\&= \frac{2x+a}{2\sqrt{x(a+x)}}- a\left(\frac{1}{2\sqrt{x(a+x)}}\right)\\ &=\frac{2x}{2\sqrt{x(a+x)}} = \frac{x}{\sqrt{x(a+x)}}\end{align*}

So,

∫xx(a+x) dx=x(a+x)−alog⁡(x+a+x)+C\begin{align*}\int \frac{x}{\sqrt{x(a+x)}}\ \dee{x}&=\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)+C\end{align*}

Remark: ∫xx(a+x) dx\int \frac{x}{\sqrt{x(a+x)}}\ \dee{x} can be calculated using the method of Trigonometric Substitution, which you will learn in Section 1.9 of the CLP-2 text.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q20Stage 2Past exam · 2016Q2

Evaluate ∫02(x3+sin⁡x) dx\displaystyle\int_0^2 \big(x^3+\sin x)\,\dee{x}.

Hint

Guess a function whose derivative is the integrand, then use the Fundamental Theorem of Calculus Part 2.

Answer

5−cos⁡25-\cos 2

Full solution

By the Fundamental Theorem of Calculus,

∫02(x3+sin⁡x) dx=[x44−cos⁡x]02=(244−cos⁡2)−(0−cos⁡0)=4−cos⁡2+1=5−cos⁡2.\begin{align*} \int_0^2 \big(x^3+\sin x)\,\dee{x} &=\left[\frac{x^4}{4}-\cos x\right]_0^2\\ &=\left(\frac{2^4}{4}-\cos 2\right)-\left(0 -\cos 0\right) \\ &= 4-\cos2+1=5-\cos2. \end{align*}
Q21Stage 2Past exam · 2012A

Evaluate ∫12x2+2x2 dx\displaystyle\int_1^2 \frac{x^2+2}{x^2}\,\dee{x}.

Hint

Split the given integral up into two integrals.

Answer

22

Full solution

By part (d) of our “Arithmetic of Integration” theorem, Theorem 1.2.1 in the CLP-2 text,

∫12x2+2x2 dx=∫12[1+2x2] dx=∫12dx+2∫121x2 dx\begin{equation*} \int_1^2 \frac{x^2+2}{x^2}\,\dee{x} =\int_1^2 \Big[1+\frac{2}{x^2}\Big]\,\dee{x} =\int_1^2 \dee{x} + 2\int_1^2 \frac{1}{x^2}\,\dee{x} \end{equation*}

Then by the Fundamental Theorem of Calculus Part 2,

∫12dx+2∫121x2 dx=[x]12+2[−1x]12=[2−1]+2[−12+1]=2\begin{align*} \int_1^2 \dee{x} + 2\int_1^2 \frac{1}{x^2}\,\dee{x} =\Big[x\Big]_1^2 + 2\Big[-\frac{1}{x}\Big]_1^2 =\big[2-1\big] + 2\Big[-\frac{1}{2}+1\Big] =2 \end{align*}
Q22Stage 2

Evaluate ∫11+25x2dx\displaystyle\int \dfrac{1}{1+25x^2}\dee{x}.

Hint

The integrand is similar to 11+x2\dfrac{1}{1+x^2}, so something with arctangent seems in order.

Answer

15arctan⁡(5x)+C\dfrac{1}{5}\arctan(5x)+C

Full solution

The integrand is similar to 11+x2\dfrac{1}{1+x^2}, which is the derivative of arctangent. Indeed, we have

∫11+25x2dx=∫11+(5x)2dx.\begin{align*}\displaystyle\int \dfrac{1}{1+25x^2}\dee{x} &= \int \dfrac{1}{1+(5x)^2}\dee{x}.\end{align*}

So, a reasonable first guess for the antiderivative might be

F(x)=?arctan⁡(5x).\begin{align*}\color{red}F(x) &\color{red}\stackrel{?}{=} \arctan(5x).\end{align*}

However, because of the chain rule,

F′(x)=51+(5x)2.\begin{align*}\color{red}F'(x) &\color{red}= \dfrac{5}{1+(5x)^2}.\end{align*}

In order to “fix" the numerator, we make a second guess:

F(x)=15arctan⁡(5x)F′(x)=15(51+(5x)2)=11+25x2So,∫11+25x2dx=15arctan⁡(5x)+C.\begin{align*}\color{blue}F(x) &\color{blue}= \frac{1}{5}\arctan(5x)\\ \color{blue}F'(x) &\color{blue}= \dfrac{1}{5}\left(\dfrac{5}{1+(5x)^2}\right) = \dfrac{1}{1+25x^2}\\ \text{So,}\qquad \displaystyle\int \dfrac{1}{1+25x^2}\dee{x}&=\frac{1}{5}\arctan(5x)+C.\end{align*}
Q23Stage 2

Evaluate ∫12−x2dx\displaystyle\int \dfrac{1}{\sqrt{2-x^2}}\dee{x}.

Hint

The integrand is similar to 11−x2\dfrac{1}{\sqrt{1-x^2}}, so factoring out 2\sqrt{2} from the denominator will make it look like some flavour of arcsine.

Answer

arcsin⁡(x2)+C\arcsin\left(\dfrac{x}{\sqrt{2}}\right)+C

Full solution

The integrand is similar to 11−x2\dfrac{1}{\sqrt{1-x^2}}. In order to formulate a guess for the antiderivative, let's factor out 2\sqrt{2} from the denominator:

∫12−x2dx=∫12(1−x22)dx=∫121−x22dx=∫12⋅11−(x2)2dx\begin{align*}\displaystyle\int \dfrac{1}{\sqrt{2-x^2}}\dee{x}&= \displaystyle\int \dfrac{1}{\sqrt{2\left(1-\frac{x^2}{2}\right)}}\dee{x}\\ &=\displaystyle\int \dfrac{1}{ \sqrt{2}\sqrt{1-\frac{x^2}{2}}}\dee{x}\\ &= \int\frac{1}{\sqrt{2}}\cdot \dfrac{1}{ \sqrt{1-\left(\dfrac{x}{\sqrt2}\right)^2}}\dee{x}\end{align*}

At this point, we might guess that our antiderivative is something like F(x)=arcsin⁡(x2)F(x) = \arcsin\left(\dfrac{x}{\sqrt{2}}\right). To explore this possibility, we can differentiate, and see what we get.

ddx{arcsin⁡(x2)}=12⋅11−(x2)2\begin{align*}\diff{}{x}\left\{\arcsin\left(\dfrac{x}{\sqrt{2}}\right)\right\}&=\frac{1}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{1-\left(\dfrac{x}{\sqrt{2}}\right)^2}}\end{align*}

This is exactly what we want! So,

∫12−x2dx=arcsin⁡(x2)+C\begin{align*}\int \dfrac{1}{\sqrt{2-x^2}}\dee{x}&=\arcsin\left(\dfrac{x}{\sqrt2}\right)+C\end{align*}
Q24Stage 2

Evaluate ∫tan⁡2x dx\displaystyle\int \tan^2 x \ \dee{x}.

Hint

We know how to antidifferentiate sec⁡2x\sec^2 x, and there is an identity linking sec⁡2x\sec^2 x with tan⁡2x\tan^2 x.

Answer

tan⁡x−x+C\tan x - x +C

Full solution

We know that ∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\ \dee{x} = \tan x +C, and sec⁡2x=tan⁡2x+1\sec^2x = \tan^2 x + 1, so

∫tan⁡2x dx=∫sec⁡2x−1 dx=∫sec⁡2x dx−∫1 dx=tan⁡x−x+C\begin{align*} \int \tan^2 x\ \dee{x} &= \int \sec^2x - 1 \ \dee{x}\\ &=\int \sec^2 x \ \dee{x} - \int 1 \ \dee{x}\\ &=\tan x - x + C \end{align*}
Q25Stage 2

Evaluate ∫3sin⁡xcos⁡x dx\displaystyle\int 3 \sin x \cos x \ \dee{x}.

Hint

Recall 2sin⁡xcos⁡x=sin⁡(2x)2\sin x \cos x = \sin(2x).

Answer

−34cos⁡(2x)+C-\dfrac{3}{4}\cos(2x)+C, or equivalently, 32sin⁡2x+C\dfrac{3}{2}\sin^2 x+C

Full solution

  • This might not obviously look like the derivative of anything familiar, but it does look like half of a familiar trig identity: 2sin⁡xcos⁡x=sin⁡(2x)2\sin x \cos x = \sin(2x).

    ∫3sin⁡xcos⁡x dx=∫32⋅2sin⁡xcos⁡x dx=∫32sin⁡(2x) dx\begin{align*}\int 3 \sin x \cos x \ \dee{x}&=\int \frac{3}{2}\cdot 2\sin x \cos x \ \dee{x}\\ &=\int \frac{3}{2} \sin(2x)\ \dee{x}\end{align*}

    So, we might guess that the antiderivative is something like −cos⁡(2x)-\cos(2x). We only need to figure out the constants.

    ddx{−cos⁡(2x)}=2sin⁡(2x)So,ddx{−34cos⁡(2x)}=32sin⁡(2x)Therefore,∫3sin⁡xcos⁡x dx=−34cos⁡(2x)+C\begin{align*}\diff{}{x}\{-\cos(2x)\}&=2\sin(2x)\\ \text{So,}\qquad \diff{}{x}\left\{-\frac{3}{4}\cos(2x)\right\}&=\frac{3}{2}\sin(2x)\\ \text{Therefore,}\qquad \int 3\sin x \cos x\ \dee{x}&=-\frac{3}{4}\cos(2x)+C\end{align*}

  • You might notice that the integrand looks like it came from the chain rule, since cos⁡x\cos x is the derivative of sin⁡x\sin x. Using this observation, we can work out the antideriative:

    ddx{sin⁡2x}=2sin⁡xcos⁡xddx{32sin⁡2x}=3sin⁡xcos⁡xSo,∫3sin⁡xcos⁡x dx=32sin⁡2x+C\begin{align*} \diff{}{x}\left\{\sin^2 x\right\}&=2\sin x \cos x\\ \diff{}{x}\left\{\frac{3}{2}\sin^2 x\right\}&=3\sin x \cos x\\ \text{So,}\qquad \int 3\sin x \cos x \ \dee{x}&=\frac{3}{2}\sin^2 x+C \end{align*}

These two answers look different. Using the identity cos⁡(2x)=1−2sin⁡2(x)\cos(2x)=1 - 2 \sin^2(x), we reconcile them:

−34cos⁡(2x)+C=−34(1−2sin⁡2x)+C=32sin⁡2x+(C−34)\begin{align*} -\frac{3}{4}\cos(2x)+C&= -\frac{3}{4}\left(1-2\sin^2x\right)+C\\ &=\frac{3}{2}\sin^2 x + \left(C-\frac{3}{4}\right) \end{align*}

The 34\frac{3}{4} here is not significant. Remember that CC is used to designate a constant that can take any value between −∞-\infty and +∞+\infty. So C−34C-\frac{3}{4} is also just a constant that can take any value between −∞-\infty and +∞+\infty. As the two answers we found differ by a constant, they are equivalent.

Q26Stage 2

Evaluate ∫cos⁡2x dx\displaystyle\int \cos^2 x \ \dee{x}.

Hint

cos⁡2x=1+cos⁡(2x)2\cos^2 x = \dfrac{1+\cos(2x)}{2}

Answer

12x+14sin⁡(2x)+C\dfrac{1}{2}x+\dfrac{1}{4}\sin(2x)+C

Full solution

It's not immediately obvious which function has cos⁡2x\cos^2 x as its derivative, but we can make the situation a little clearer by using the identity cos⁡2x=1+cos⁡(2x)2\cos^2 x = \dfrac{1+\cos(2x)}{2}:

∫cos⁡2x dx=∫12⋅(1+cos⁡(2x)) dx=∫12 dx+∫12cos⁡(2x) dx=12x+C+∫12cos⁡(2x) dx\begin{align*}\int \cos^2 x \ \dee{x}&=\int \frac{1}{2}\cdot \left(1+\cos(2x)\right)\ \dee{x}\\ &=\int \frac{1}{2}\ \dee{x} + \int \frac{1}{2}\cos(2x)\ \dee{x}\\ &=\frac{1}{2}x + C+ \int \frac{1}{2}\cos(2x)\ \dee{x}\end{align*}

For the remaining integral, we might guess something like F(x)=sin⁡(2x)F(x) = \sin(2x). Let's figure out the appropriate constant:

ddx{sin⁡(2x)}=2cos⁡(2x)ddx{14sin⁡(2x)}=12cos⁡(2x)So,∫12cos⁡(2x) dx=14sin⁡(2x)+CTherefore,∫cos⁡2x dx=12x+14sin⁡(2x)+C\begin{align*}\diff{}{x}\left\{ \sin(2x) \right\}&=2\cos(2x)\\ \diff{}{x}\left\{ \frac{1}{4}\sin(2x) \right\}&=\frac{1}{2}\cos(2x)\\ \text{So,}\qquad \int \frac{1}{2}\cos(2x)\ \dee{x} &=\frac{1}{4}\sin(2x)+C\\ \text{Therefore,}\qquad \int \cos^2 x \ \dee{x}&=\frac{1}{2}x+\frac{1}{4}\sin(2x)+C\end{align*}
Q27Stage 2Past exam · M105 2012A

If

F(x)=∫0xlog⁡(2+sin⁡t) dtandG(y)=∫y0log⁡(2+sin⁡t) dt\begin{equation*} F(x)=\int_0^x \log(2+\sin t)\,\dee{t}\quad\text{and}\quad G(y)=\int^0_y \log(2+\sin t)\,\dee{t} \end{equation*}

find F′(π2)F'\big(\frac{\pi}{2}\big) and G′(π2)G'\big(\frac{\pi}{2}\big).

Answer

F′(π2)=log⁡(3)F'\left(\frac{\pi}{2}\right)=\log(3) G′(π2)=−log⁡(3)G'\left(\frac{\pi}{2}\right)=-\log(3)

Full solution

By the Fundamental Theorem of Calculus Part 1,

F′(x)=ddx∫0xlog⁡(2+sin⁡t) dt=log⁡(2+sin⁡x)G′(y)=ddy[−∫0ylog⁡(2+sin⁡t) dt]=−log⁡(2+sin⁡y)\begin{alignat*}{3} F'(x)&=\diff{}{x} \int_0^x \log(2+\sin t)\,\dee{t} &&=\log(2+\sin x) \\ G'(y)&=\diff{}{y}\bigg[- \int_0^y \log(2+\sin t)\,\dee{t}\bigg] &&=-\log(2+\sin y) \\ \end{alignat*}

So,

F′(π2)=log⁡3G′(π2)=−log⁡(3)\begin{equation*} F'\Big(\frac{\pi}{2}\Big)=\log 3\qquad G'\Big(\frac{\pi}{2}\Big)=-\log(3) \end{equation*}
Q28Stage 2Past exam · 2014A

Let f(x)=∫1x100(t2−3t+2)e−t2 dtf(x)=\displaystyle\int_1^x 100(t^2-3t+2)e^{-t^2}\ \dee{t}. Find the interval(s) on which ff is increasing.

Hint

There is a good way to test where a function is increasing, decreasing, or constant, that also has something to do with topic of this section.

Answer

f(x)f(x) is increasing when −∞<x<1-\infty<x<1 and when 2<x<∞2<x<\infty.

Full solution

By the Fundamental Theorem of Calculus Part 1,

f′(x)=100(x2−3x+2)e−x2=100(x−1)(x−2)e−x2\begin{equation*} f'(x) = 100(x^2-3x+2)e^{-x^2} = 100(x-1)(x-2)e^{-x^2} \end{equation*}

As f(x)f(x) is increasing whenever f′(x)>0f'(x)>0 and 100e−x2100 e^{-x^2} is always strictly bigger than 00, we have f(x)f(x) increasing if and only if (x−1)(x−2)>0(x-1)(x-2)>0, which is the case if and only if (x−1)(x-1) and (x−2)(x-2) are of the same sign. Both are positive when x>2x>2 and both are negative when x<1x<1. So f(x)f(x) is increasing when −∞<x<1-\infty<x<1 and when 2<x<∞2<x<\infty.

Remark: even without the Fundamental Theorem of Calculus, since f(x)f(x) is the area under a curve from 1 to xx, f(x)f(x) is increasing when the curve is above the xx-axis (because we're adding positive area), and it's decreasing when the curve is below the xx-axis (because we're adding negative area).

Q29Stage 2Past exam · M105 2013A

If F(x)=∫0cos⁡x1t3+6 dtF(x)={\displaystyle\int_0^{\cos x} \frac{1}{t^3+6}\,\dee{t}}, find F′(x)F'(x).

Hint

See Example 1.3.5 in the

CLP-2 text.

Answer

F′(x)=−sin⁡xcos⁡3x+6F'(x)=-\dfrac{\sin x}{\cos^3x+6}

Full solution

Write G(x)=∫0x1t3+6 dtG(x)={\displaystyle\int_0^x \frac{1}{t^3+6}\,\dee{t}}. By the Fundamental Theorem of Calculus Part 1, G′(x)=1x3+6G'(x)=\dfrac{1}{x^3+6}. Since F(x)=G(cos⁡x)F(x)=G(\cos x), the chain rule gives us

F′(x)=G′(cos⁡x)⋅(−sin⁡x)=−sin⁡xcos⁡3x+6\begin{align*} F'(x)=G'(\cos x)\cdot(-\sin x)=-\frac{\sin x}{\cos^3x+6} \end{align*}
Q30Stage 2Past exam · 1997D

Compute f′(x)f'(x) where f(x)=∫01+x4et2dtf(x)= \displaystyle\int_0^{1+x^4}e^{t^2}\dee{t}.

Hint

See Example 1.3.5 in the

CLP-2 text.

Answer

4x3e(1+x4)24x^3e^{(1+x^4)^2}

Full solution

Define g(x)=∫0xet2dtg(x)= \displaystyle\int_0^x e^{t^2}\dee{t}. By the Fundamental Theorem of Calculus Part 1, g′(x)=ex2g'(x)=e^{x^2}. As f(x)=g(1+x4)f(x)=g(1+x^4) the chain rule gives us

f′(x)=4x3g′(1+x4)=4x3e(1+x4)2\begin{align*} f'(x)=4x^3g'(1+x^4)=4x^3e^{(1+x^4)^2} \end{align*}
Q31Stage 2Past exam · M105 2015A

Evaluate ddx{∫0sin⁡x(t6+8)dt}\displaystyle\diff{}{x}\left\{\int_0^{\sin x}(t^6+8)\dee{t}\right\}.

Hint

See Example 1.3.5 in the

CLP-2 text.

Answer

(sin⁡6x+8)cos⁡x\big(\sin^6 x+8)\cos x

Full solution

Define g(x)=∫0x(t6+8)dtg(x)=\int_0^x (t^6+8)\dee{t}. By the fundamental theorem of calculus, g′(x)=x6+8g'(x)=x^6+8. We are to compute the derivative of f(x)=g(sin⁡x)f(x)=g(\sin x). The chain rule gives

ddx{∫0sin⁡x(t6+8)dt}=g′(sin⁡x)⋅cos⁡x=(sin⁡6x+8)cos⁡x\begin{align*} \diff{}{x}\left\{\int_0^{\sin x}(t^6+8)\dee{t}\right\} =g'(\sin x)\cdot \cos x=\big(\sin^6 x+8\big)\cos x \end{align*}
Q32Stage 2Past exam · 2000D

Let F(x)=∫0x3e−tsin⁡(πt2) dtF(x)= \displaystyle\int_0^{x^3}e^{-t}\sin\left(\frac{\pi t}{2}\right)\,\dee{t}. Calculate F′(1)F'(1).

Hint

See Example 1.3.5 in the

CLP-2 text.

Answer

F′(1)=3e−1F'(1)=3e^{-1}

Full solution

Let G(x)=∫0xe−tsin⁡(πt2) dtG(x)= \displaystyle\int_0^{x}e^{-t}\sin\left(\frac{\pi t}{2}\right)\,\dee{t}. By the Fundamental Theorem of Calculus Part 1, G′(x)=e−xsin⁡(πx2)G'(x)=e^{-x}\sin\big(\frac{\pi x}{2}\big) and, since F(x)=G(x3)F(x)=G(x^3), F′(x)=3x2G′(x3)=3x2e−x3sin⁡(πx32)F'(x)=3x^2G'(x^3)=3x^2e^{-x^3}\sin\big(\frac{\pi x^3}{2}\big). Then F′(1)=3e−1sin⁡(π2)=3e−1F'(1)=3e^{-1}\sin\big(\frac{\pi }{2}\big) =3e^{-1}.

Q33Stage 2Past exam · 2016Q2

Find ddu{∫cos⁡u0dt1+t3}\displaystyle \diff{}{u} \left\{ \int_{\cos u}^0 \frac{\dee{t}}{1+t^3} \right\}.

Hint

See Example 1.3.6 in the

CLP-2 text.

Answer

sin⁡u1+cos⁡3u\displaystyle{}\frac{\sin{u}}{1+\cos^3 u}

Full solution

Define G(x)=∫x0dt1+t3=−∫0x11+t3 dt\displaystyle G(x) = \int_x^0 \frac{\dee{t}}{1+t^3} = - \int_0^x \frac1{1+t^3}\,\dee{t}, so that G′(x)=−11+x3\displaystyle G'(x) = - \frac1{1+x^3} by the Fundamental Theorem of Calculus Part 1. Then by the chain rule,

ddu{∫cos⁡u0dt1+t3}=dduG(cos⁡u)=G′(cos⁡u)⋅dducos⁡u=−11+cos⁡3u⋅(−sin⁡u).\begin{align*} \diff{}{u} \left\{ \int_{\cos u}^0 \frac{\dee{t}}{1+t^3} \right\} = \diff{}{u} G(\cos u) = G'(\cos u) \cdot \diff{}{u}\cos u = {-}\frac{1}{1+\cos^3 u} \cdot (-\sin{u}). \end{align*}
Q34Stage 2Past exam · M121 2000A

Find f(x)f(x) if x2=1+∫1xf(t) dtx^2=1+\displaystyle\int_1^x f(t)\ \dee{t}.

Hint

Apply ddx\diff{}{x} to both sides.

Answer

f(x)=2xf(x)=2x

Full solution

Applying ddx\diff{}{x} to both sides of x2=1+∫1xf(t) dtx^2=1+\int_1^x f(t)\ \dee{t} gives, by the Fundamental Theorem of Calculus Part 1, 2x=f(x)2x=f(x).

Q35Stage 2Past exam · 2013A

If xsin⁡(πx)=∫0xf(t) dtx \sin(\pi x) = \displaystyle\int_0^x f(t)\, \dee{t} where ff is a continuous function, find f(4)f(4).

Hint

What is the title of this section?

Answer

f(4)=4πf(4)=4\pi

Full solution

Apply ddx\diff{}{x} to both sides of xsin⁡(πx)=∫0xf(t) dtx \sin(\pi x) = \int_0^x f(t)\, \dee{t}. Then, by the Fundamental Theorem of Calculus Part 1,

f(x)=ddx∫0xf(t) dt=ddx{xsin⁡(πx)}  ⟹  f(x)=ddx{xsin⁡(πx)}=sin⁡(πx)+πxcos⁡(πx)  ⟹  f(4)=sin⁡(4π)+4πcos⁡(4π)=4π\begin{alignat*}{3} &&f(x) =\diff{}{x} \int_0^x f(t)\, \dee{t}&=\diff{}{x}\big\{x \sin(\pi x)\big\} \\ \implies&& f(x) &= \diff{}{x}\big\{x \sin(\pi x)\big\} =\sin(\pi x)+\pi x\cos(\pi x) \\ \implies&& f(4)&= \sin(4\pi)+4\pi\cos(4\pi)=4\pi \end{alignat*}
Q36Stage 2Past exam · 2016Q2

Consider the function F(x)=∫0x2e−t dt+∫−x0e−t2 dt\displaystyle F(x)=\int_0^{x^2} e^{-t}\,\dee{t} +\int_{-x}^0 e^{-t^2}\,\dee{t}.

  1. Find F′(x)F'(x).

  2. Find the value of xx for which F(x)F(x) takes its minimum value.

Hint

See Example 1.3.6 in the

CLP-2 text.

Answer

(a) (2x+1)e−x2(2x+1)e^{-x^2} (b) x=−1/2x=-1/2

Full solution

(a) Write

F(x)=G(x2)−H(−x)withG(y)=∫0ye−t dt, H(y)=∫0ye−t2 dt\begin{align*} F(x)=G(x^2)-H(-x)\qquad\hbox{with}\quad G(y)=\int_0^{y} e^{-t}\,\dee{t},\ H(y)=\int^{y}_0 e^{-t^2}\,\dee{t} \end{align*}

By the Fundamental Theorem of Calculus Part 1,

G′(y)=e−y ,H′(y)=e−y2\begin{align*} G'(y)=e^{-y}\, ,\qquad H'(y)= e^{-y^2} \end{align*}

Hence, by the chain rule,

F′(x)=2xG′(x2)−(−1)H′(−x)=2xe−(x2)+e−(−x)2=(2x+1)e−x2\begin{align*} F'(x)=2x G'(x^2)-(-1)H'(-x)=2x e^{-(x^2)}+e^{-(-x)^2} =(2x+1)e^{-x^2} \end{align*}

(b) Observe that F′(x)<0F'(x)<0 for x<−1/2x<-1/2 and F′(x)>0F'(x)>0 for x>−1/2x>-1/2. Hence F(x)F(x) is decreasing for x<−1/2x<-1/2 and increasing for x>−1/2x>-1/2, and F(x)F(x) must take its minimum value when x=−1/2x=-1/2.

Q37Stage 2Past exam · 2016A

If F(x)F(x) is defined by F(x)=∫x4−x3xesin⁡t dt\displaystyle F(x) = \int_{x^4-x^3}^x e^{\sin t}\,\dee{t}, find F′(x)F'(x).

Hint

See Example 1.3.6 in the CLP-2 text.

Answer

esin⁡x−esin⁡(x4−x3)(4x3−3x2)e^{\sin x}-e^{\sin(x^4-x^3)}\big(4x^3-3x^2\big)

Full solution

Define G(y)=∫0yesin⁡t dtG(y)=\displaystyle\int_0^ye^{\sin t}\ \dee{t}. Then:

F(x)=∫0xesin⁡t dt+∫x4−x30esin⁡t dt=∫0xesin⁡t dt−∫0x4−x3esin⁡t dt=G(x)−G(x4−x3)\begin{align*} F(x) &= \int_0^x e^{\sin t}\,\dee{t} + \int_{x^4-x^3}^0 e^{\sin t}\,\dee{t} = \int_0^x e^{\sin t}\,\dee{t} - \int_0^{x^4-x^3} e^{\sin t}\,\dee{t} \\ &=G(x) - G(x^4-x^3) \end{align*}

By the Fundamental Theorem of Calculus Part 1,

G′(y)=esin⁡y\begin{align*} G'(y)=e^{\sin y}\qquad \end{align*}

Hence, by the chain rule,

F′(x)=G′(x)−G′(x4−x3) ddx{x4−x3}=G′(x)−G′(x4−x3) (4x3−3x2)=esin⁡x−esin⁡(x4−x3)(4x3−3x2)\begin{align*} F'(x) &= G'(x)- G'(x^4-x^3)\ \diff{}{x}\big\{x^4-x^3\big\} \\ &= G'(x)- G'(x^4-x^3)\ (4x^3-3x^2) \\ &=e^{\sin x}-e^{\sin(x^4-x^3)}\big(4x^3-3x^2\big) \end{align*}
Q38Stage 2Past exam · M121 2012A

Evaluate ddx{∫x5−x2cos⁡(et) dt\displaystyle \diff{}{x}\bigg\{\int_{x^5}^{-x^2} \cos\big(e^t\big)\,\dee{t}}.

Hint

See Example 1.3.6 in the CLP-2 text.

Answer

−2xcos⁡(e−x2)−5x4cos⁡(ex5)-2x \cos\big(e^{-x^2}\big) -5x^4\cos\big(e^{x^5}\big)

Full solution

Define with G(y)=∫0ycos⁡(et) dtG(y) = \displaystyle\int_0^y \cos\big(e^t\big)\,\dee{t}. Then:

F(x)=∫x5−x2cos⁡(et) dt=∫0−x2cos⁡(et) dt+∫x50cos⁡(et) dt=∫0−x2cos⁡(et) dt−∫0x5cos⁡(et) dt=G(−x2)−G(x5)\begin{align*} F(x) &= \int_{x^5}^{-x^2} \cos\big(e^t\big)\,\dee{t} =\int_0^{-x^2} \cos\big(e^t\big)\,\dee{t} + \int_{x^5}^0 \cos\big(e^t\big)\,\dee{t} \\ &=\int_0^{-x^2} \cos\big(e^t\big)\,\dee{t} - \int_0^{x^5} \cos\big(e^t\big)\,\dee{t} \\ &=G(-x^2) - G(x^5) \end{align*}

By the Fundamental Theorem of Calculus,

G′(y)=cos⁡(ey)\begin{align*} G'(y)=\cos\big(e^y\big) \end{align*}

Hence, by the chain rule,

F′(x)=G′(−x2) ddx{−x2}−G′(x5) ddx{x5}=G′(−x2) (−2x)−G′(x5) (5x4)=−2xcos⁡(e−x2)−5x4cos⁡(ex5)\begin{align*} F'(x) &= G'(-x^2)\ \diff{}{x}\big\{-x^2\big\} - G'(x^5)\ \diff{}{x}\big\{x^5\big\} \\ &= G'(-x^2)\ (-2x)- G'(x^5)\ (5x^4) \\ &=-2x \cos\big(e^{-x^2}\big) -5x^4\cos\big(e^{x^5}\big) \end{align*}
Q39Stage 2Past exam · 2014D

Differentiate ∫xexsin⁡t dt\displaystyle \int_x^{e^x} \sqrt{\sin t}\,\dee{t} for 0<x<log⁡π0<x<\log \pi.

Hint

See Example 1.3.6 in the CLP-2 text.

Answer

exsin⁡(ex)−sin⁡(x)e^x\sqrt{\sin(e^x)} -\sqrt{\sin(x)}

Full solution

Define with G(y)=∫0ysin⁡t dtG(y) = \displaystyle\int_0^y \sqrt{\sin t}\,\dee{t}. Then:

F(x)=∫xexsin⁡t dt=∫0exsin⁡t dt+∫x0sin⁡t dt=∫0exsin⁡t dt−∫0xsin⁡t dt=G(ex)−G(x)\begin{align*} F(x)&=\int_x^{e^x} \sqrt{\sin t}\,\dee{t} \\ &= \int_0^{e^x} \sqrt{\sin t}\,\dee{t} + \int_x^0 \sqrt{\sin t}\,\dee{t} = \int_0^{e^x} \sqrt{\sin t}\,\dee{t} - \int_0^x \sqrt{\sin t}\,\dee{t} \\ &=G(e^x) - G(x) \end{align*}

By the Fundamental Theorem of Calculus Part 1,

G′(y)=sin⁡y\begin{align*} G'(y)=\sqrt{\sin y}\qquad \end{align*}

Hence, by the chain rule,

F′(x)=G′(ex) ddx{ex}−G′(x)=exG′(ex)−G′(x)=exsin⁡(ex)−sin⁡(x)\begin{align*} F'(x) &= G'(e^x)\ \diff{}{x}\big\{e^x\big\} - G'(x)\\ &= e^xG'(e^x) - G'(x)\\ &=e^x\sqrt{\sin(e^x)} -\sqrt{\sin(x)} \end{align*}
Q40Stage 2Past exam · M105 2014A

Evaluate ∫15f(x) dx\displaystyle \int_1^5 f(x)\,\dee{x}, where f(x)={3 if x≤3x if x≥3\displaystyle f(x)= \begin{cases} 3 &\text{ if }x\le 3 \\ x &\text{ if }x\ge 3 \end{cases}.

Hint

Split up the domain of integration.

Answer

1414

Full solution

Splitting up the domain of integration,

∫15f(x) dx=∫13f(x) dx+∫35f(x) dx=∫133 dx+∫35x dx=3x∣x=1x=3+x22∣x=3x=5=14\begin{align*} \int_1^5 f(x)\ \dee{x} &= \int_1^3 f(x)\,\dee{x} + \int_3^5 f(x)\,\dee{x} \\ &= \int_1^3 3\,\dee{x} + \int_3^5 x\,\dee{x} \\ &= 3x\bigg|_{x=1}^{x=3} +\frac{x^2}{2} \bigg|_{x=3}^{x=5} \\ &=14 \end{align*}

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q41Stage 3Past exam · M105 2014A

If f′(1)=2f'(1)=2 and f′(2)=3f'(2)=3, find ∫12f′(x)f′′(x) dx\displaystyle\int_1^2 f'(x) f''(x)\,\dee{x}.

Hint

It is possible to guess an antiderivative for f′(x)f′′(x)f'(x) f''(x) that is expressed in terms of f′(x)f'(x).

Answer

52\dfrac{5}{2}

Full solution

By the chain rule,

ddx{(f′(x))2}=2f′(x) f′′(x)\begin{align*} \diff{}{x}\big\{\left(f'(x)\right)^2\big\} = 2 f'(x)\,f''(x) \end{align*}

so 12f′(x)2\frac{1}{2} f'(x)^2 is an antiderivative for f′(x) f′′(x)f'(x)\,f''(x) and, by the Fundamental Theorem of Calculus Part 2,

∫12f′(x)f′′(x) dx=[12(f′(x))2]x=1x=2=12f′(2)2−12f′(1)2=52\begin{align*} \int_1^2 f'(x) f''(x)\,\dee{x} =\left[\frac{1}{2}\left(f'(x)\right)^2\right]_{x=1}^{x=2} =\frac{1}{2} f'(2)^2 - \frac{1}{2} f'(1)^2 =\frac{5}{2} \end{align*}

Remark: evaluating antiderivatives of this type will occupy the next section, Section 1.4 of the CLP-2 text.

Q42Stage 3Past exam · 2016Q2

A car traveling at 30 m/s30\,\textrm{m}/\textrm{s} applies its brakes at time t=0t=0, its velocity (in m/s\textrm{m}/\textrm{s}) decreasing according to the formula v(t)=30−10tv(t) = 30 - 10t. How far does the car go before it stops?

Hint

When does the car stop? What is the relation between velocity and distance travelled?

Answer

45 m45\,\textrm{m}

Full solution

The car stops when v(t)=30−10t=0v(t)=30-10t=0, which occurs at time t=3t = 3. The distance covered up to that time is

∫03v(t) dt=(30t−5t2)∣03=(90−45)−0=45 m.\begin{align*} \int_0^3 v(t)\,\dee{t} = (30t - 5t^2)\Big|_{0}^{3} = (90-45)-0 = 45\,\textrm{m}. \end{align*}
Q43Stage 3Past exam · 1998A

Compute f′(x)f'(x) where f(x)=∫02x−x2log⁡(1+et) dtf(x)= \displaystyle\int_0^{2x-x^2}\log\big(1+e^t\big)\,\dee{t}. Does f(x)f(x) have an absolute maximum? Explain.

Hint

See Example 1.3.5 in the

CLP-2 text. For the absolute maximum part of the question, study the sign of f′(x)f'(x).

Answer

f′(x)=(2−2x)log⁡(1+e2x−x2)f'(x)=(2-2x)\log\big(1+e^{2x-x^2}\big) and f(x)f(x) achieves its absolute maximum at x=1x=1, because f(x)f(x) is increasing for x<1x<1 and decreasing for x>1x>1.

Full solution

Define g(x)=∫0xlog⁡(1+et) dtg(x) = \displaystyle\int_0^x\log\big(1+e^t\big)\,\dee{t}. By the Fundamental Theorem of Calculus Part 1, g′(x)=log⁡(1+ex)g'(x) = \log\big(1+e^x\big). But f(x)=g(2x−x2)f(x)=g(2x-x^2), so by the chain rule,

f′(x)=g′(2x−x2)⋅ddx{2x−x2}=(2−2x)⋅log⁡(1+e2x−x2)\begin{align*} f'(x)=g'(2x-x^2)\cdot \diff{}{x}\{2x-x^2\} =(2-2x)\cdot\log\big(1+e^{2x-x^2}\big) \end{align*}

Observe that e2x−x2>0e^{2x-x^2}>0 for all xx so that 1+e2x−x2>11+e^{2x-x^2}>1 for all xx and log⁡(1+e2x−x2)>0\log\big(1+e^{2x-x^2}\big)>0 for all xx. Since 2−2x2-2x is positive for x<1x<1 and negative for x>1x>1, f′(x)f'(x) is also positive for x<1x<1 and negative for x>1x>1. That is, f(x)f(x) is increasing for x<1x<1 and decreasing for x>1x>1. So f(x)f(x) achieves its absolute maximum at x=1x=1.

Q44Stage 3Past exam · 2001A

Find the minimum value of ∫0x2−2xdt1+t4\displaystyle\int_0^{x^2-2x}\frac{\dee{t}}{1+t^4}. Express your answer as an integral.

Hint

See Example 1.3.5 in the

CLP-2 text. For the “minimum value” part of the question, study the sign of f′(x)f'(x).

Answer

The minimum is ∫0−1dt1+t4\int_0^{-1} \frac{\dee{t}}{1+t^4}. As xx runs from −∞-\infty to ∞\infty, the function f(x)=∫0x2−2xdt1+t4f(x)= \int_0^{x^2-2x}\frac{\dee{t}}{1+t^4} decreases until xx reaches 1 and then increases all x>1x>1. So the minimum is achieved for x=1x=1. At x=1x=1, x2−2x=−1x^2-2x=-1.

Full solution

Let f(x)=∫0x2−2xdt1+t4f(x)=\int_0^{x^2-2x}\frac{\dee{t}}{1+t^4} and g(x)=∫0xdt1+t4g(x)= \int_0^{x}\frac{\dee{t}}{1+t^4}. Then g′(x)=11+x4g'(x)=\frac{1}{1+x^4} and, since f(x)=g(x2−2x)f(x)=g(x^2-2x), f′(x)=(2x−2)g′(x2−2x)=2x−11+(x2−2x)4f'(x)=(2x-2)g'(x^2-2x)=2\frac{x-1}{1+(x^2-2x)^4}. This is zero for x=1x=1, negative for x<1x<1 and positive for x>1x>1. Thus as xx runs from −∞-\infty to ∞\infty, f(x)f(x) decreases until xx reaches 1 and then increases all x>1x>1. So the minimum of f(x)f(x) is achieved for x=1x=1. At x=1x=1, x2−2x=−1x^2-2x=-1 and f(1)=∫0−1dt1+t4f(1)=\int_0^{-1}\frac{\dee{t}}{1+t^4}.

Q45Stage 3Past exam · 2001D

Define the function F(x)=∫0x2sin⁡(t) dtF(x)=\displaystyle\int_0^{x^2}\sin(\sqrt{t})\,\dee{t} on the interval 0<x<40<x<4. On this interval, where does F(x)F(x) have a maximum?

Hint

See Example 1.3.5 in the

CLP-2 text. For the “maximum” part of the question, study the sign of F′(x)F'(x).

Answer

FF achieves its maximum value at x=πx=\pi.

Full solution

Define G(x)=∫0xsin⁡(t) dtG(x)=\displaystyle\int_0^x\sin(\sqrt{t})\,\dee{t}. By the Fundamental Theorem of Calculus Part 1, G′(x)=sin⁡(x)G'(x)=\sin(\sqrt{x}). Since F(x)=G(x2)F(x)=G(x^2), and since x>0x>0, we have

F′(x)=2xG′(x2)=2xsin⁡∣x∣=2xsin⁡x.F'(x)=2xG'(x^2)=2x\sin |x|=2x\sin x.

Thus FF increases as xx runs from to 00 to π\pi (since F′(x)>0F'(x)>0 there) and decreases as xx runs from π\pi to 44 (since F′(x)<0F'(x)<0 there). Thus FF achieves its maximum value at x=πx=\pi.

Q46Stage 3Past exam · 2002A

Evaluate lim⁡n→∞πn∑j=1nsin⁡(jπn)\lim\limits_{n\rightarrow\infty}\dfrac{\pi}{n}\displaystyle\sum\limits_{j=1}^n \sin\left(\frac{j\pi}{n}\right) by interpreting it as a limit of Riemann sums.

Hint

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the

CLP-2 text.

Answer

22

Full solution

The given sum is of the form

lim⁡n→∞∑j=1nπnsin⁡(jπn)=lim⁡n→∞∑j=1nf(xj∗)Δx\begin{align*} \lim_{n\rightarrow\infty}\sum_{j=1}^n \frac{\pi}{n}\sin\Big(\frac{j\pi}{n}\Big) =\lim_{n\rightarrow\infty}\sum_{j=1}^n f(x_j^*)\De x \end{align*}

with Δx=πn\De x=\frac{\pi}{n}, xj∗=jπnx_j^*=\frac{j\pi}{n} and f(x)=sin⁡(x)f(x)=\sin(x). Since x0∗=0x_0^*=0 and xn∗=πx_n^*=\pi, the right hand side is the definition (using the right Riemann sum) of

∫0πf(x) dx=∫0πsin⁡(x) dx=[−cos⁡(x)]0π=2\begin{align*} \int_0^\pi f(x)\,\dee{x}=\int_0^\pi \sin(x)\,\dee{x} =\left[-\cos(x)\right]_0^\pi=2 \end{align*}

where we evaluate the definite integral using the Fundamental Theorem of Calculus Part 2.

Q47Stage 3Past exam · M121 2002A

Use Riemann sums to evaluate the limit lim⁡n→∞1n∑j=1n11+jn\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n} \sum_{j=1}^n \frac{1}{1+\frac{j}{n}} .

Hint

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the CLP-2 text.

Answer

log⁡2\log 2

Full solution

The given sum is of the form

lim⁡n→∞1n∑j=1n11+jn=lim⁡n→∞∑j=1nf(xj)Δx\begin{equation*} \lim_{n\rightarrow\infty}\frac{1}{n} \sum_{j=1}^n \frac{1}{1+\frac{j}{n}} =\lim_{n\rightarrow\infty}\sum_{j=1}^n f(x_j)\De x \end{equation*}

with Δx=1n\De x=\frac{1}{n}, xj=jnx_j=\frac{j}{n} and f(x)=11+xf(x)=\frac{1}{1+x}. The right hand side is the definition (using the right Riemann sum) of

∫01f(x) dx=∫0111+x dx=log⁡∣1+x∣∣01=log⁡2\begin{align*} \int_0^1 f(x)\,\dee{x}=\int_0^1 \frac{1}{1+x}\,\dee{x} =\log|1+x|\Big|_0^1 =\log 2 \end{align*}
Q48Stage 3

Below is the graph of y=f(t)y=f(t), −5≤t≤5-5 \leq t \leq 5. Define F(x)=∫0xf(t) dtF(x) = \displaystyle\int_{0}^x f(t)\ \dee{t} for any xx in [−5,5][-5,5]. Sketch F(x)F(x).

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Hint

Carefully check the Fundamental Theorem of Calculus: as written, it only applies directly to F(x)F(x) when x≥0x\ge0.

Is F(x)F(x) even or odd?

Answer

In the sketch below, open dots denote inflection points, and closed dots denote extrema.

Figure from prob_s1.3, line 2

Figure from prob_s1.3, line 2

Full solution

  • We learned quite a lot last semester about curve sketching. We can use those techniques here. We have to be quite careful about the sign of xx, though. We can only directly apply the Fundamental Theorem of Calculus Part 1 (as it's written in your text) when x≥0x\ge 0. So first, let's graph the right-hand portion. Notice f(x)f(x) has even symmetry–so, if we know one half of F(x)F(x), we should be able to figure out the other half with relative ease.

    • F(0)=∫00f(t) dt=0F(0)=\displaystyle\int_0^0 f(t)\ \dee{t}=0 (so, F(x)F(x) passes through the origin)

    • Using the Fundamental Theorem of Calculus Part 1, F′(x)>0F'(x)>0 when 0<x<10<x<1 and when 3<x<53<x<5; F′(x)<0F'(x)<0 when 1<x<31<x<3. So, F(x)F(x) is decreasing from 1 to 3, and increasing from 0 to 1 and also from 3 to 5. That gives us a skeleton to work with.

      Figure from prob_s1.3, line 2

      Figure from prob_s1.3, line 2

      We get the relative sizes of the maxes and mins by eyeballing the area under y=f(t)y=f(t). The first lobe (from x=0x=0 to x=1x=1 has a small positive area, so F(1)F(1) is a small positive number. The next lobe (from x=1x=1 to x=3x=3) has a larger absolute area than the first, so F(3)F(3) is negative. Indeed, the second lobe seems to have more than twice the area of the first, so ∣F(3)∣|F(3)| should be larger than F(1)F(1). The third lobe is larger still, and even after subtracting the area of the second lobe it looks much larger than the first or second lobe, so ∣F(3)∣<F(5)|F(3)|<F(5).

    • We can use F′′(x)F''(x) to get the concavity of F(x)F(x). Note F′′(x)=f′(x)F''(x)=f'(x). We observe f(x)f(x) is decreasing on (roughly) (0,2.5)(0,2.5) and (4,5)(4,5), so F(x)F(x) is concave down on those intervals. Further, f(x)f(x) is increasing on (roughly) (2.5,4)(2.5,4), so F(x)F(x) is concave up there, and has inflection points at about x=2.5x=2.5 and x=4x=4.

      Figure from prob_s1.3, line 2

      Figure from prob_s1.3, line 2

      In the sketch above, closed dots are extrema, and open dots are inflection points.

  • Now we can consider the left half of the graph. If you stare at it long enough, you might convince yourself that F(x)F(x) is an odd function. We can also show this with the following calculation:

    F(−x)=∫0−xf(t) dtAs in Example 1.2.9 of the CLP-2 text, since f(t) is even,=∫x0f(t) dt=−∫0xf(t) dt=−F(x)\begin{alignat*}{3} F(-x)&=\int_0^{-x} f(t)\ \dee{t} &&\qquad\text{As in Example~\text{1.2.9} of the CLP-2 text, since }f(t)\text{ is even,}\\ &=\int_x^0 f(t)\ \dee{t}=-\int_0^x f(t)\ \dee{t}\\ &=-F(x) \end{alignat*}

    Knowing that F(x)F(x) is odd allows us to finish our sketch.

    Figure from prob_s1.3, line 2

    Figure from prob_s1.3, line 2

Q49Stage 3Past exam · 2015A

Define f(x)=x3∫0x3+1et3dtf(x)=x^3\displaystyle\int_{0}^{x^3+1} e^{t^3} \dee{t}.

  1. Find a formula for the derivative f′(x)f'(x). (Your formula may include an integral sign.)

  2. Find the equation of the tangent line to the graph of y=f(x)y=f(x) at x=−1x=-1.

Hint

In general, the equation of the tangent line to the graph of y=f(x)y=f(x) at x=ax=a is y=f(a)+f′(a) (x−a)y=f(a) + f'(a)\,(x-a).

Answer

(a) 3x2∫0x3+1et3dt+3x5e(x3+1)33x^2 \displaystyle\int_{0}^{x^3+1} e^{t^3} \dee{t} + 3x^5 e^{(x^3+1)^3} (b) y=−3(x+1)y = -3(x+1)

Full solution

(a) Using the product rule, followed by the chain rule, followed by the Fundamental Theorem of Calculus Part 1,

f′(x)=3x2∫0x3+1et3dt+x3ddx∫0x3+1et3dt=3x2∫0x3+1et3dt+x3 [3x2][ddy∫0yet3dt]y=x3+1=3x2∫0x3+1et3dt+x3 [3x2][ey3]y=x3+1=3x2∫0x3+1et3dt+x3 [3x2]e(x3+1)3=3x2∫0x3+1et3dt+3x5e(x3+1)3\begin{align*} f'(x) & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\diff{}{x}\int_{0}^{x^3+1} e^{t^3} \dee{t} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\ \big[3x^2\big] \Bigg[\frac{d}{\dee{y}}\int_{0}^{y} e^{t^3} \dee{t}\Bigg]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\ \big[3x^2\big] \Big[e^{y^3}\Big]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\ \big[3x^2\big] e^{(x^3+1)^3} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + 3x^5 e^{(x^3+1)^3} \end{align*}

(b) In general, the equation of the tangent line to the graph of y=f(x)y=f(x) at x=ax=a is

y=f(a)+f′(a) (x−a)\begin{equation*} y=f(a) + f'(a)\,(x-a) \end{equation*}

Substituting in the given f(x)f(x) and a=−1a=-1:

f(a)=f(−1)=(−1)3∫00et3 dt=0f′(a)=f′(−1)=3(−1)2∫00et3 dt+3(−1)5e0=0−3=−3(x−a)=x−(−1)=x+1\begin{align*}f(a)=f(-1)&=(-1)^3\int_0^0 e^{t^3}\ \dee{t}=0\\ f'(a)=f'(-1)&=3(-1)^2\int_{0}^{0}e^{t^3}\ \dee{t} + 3(-1)^5e^{0}\\ &=0-3=-3\\ (x-a)=x-(-1)&=x+1\end{align*}

So, the equation of the tangent line is

y=−3(x+1) .\begin{align*}y = -3(x+1)\ .\end{align*}
Q50Stage 3

Two students calculate ∫f(x) dx\int f(x)\ \dee{x} for some function f(x)f(x).

  • Student A calculates ∫f(x) dx=tan⁡2x+x+C\int f(x)\ \dee{x} = \tan^2 x + x + C

  • Student B calculates ∫f(x) dx=sec⁡2x+x+C\int f(x)\ \dee{x} = \sec^2 x + x + C

  • It is a fact that ddx{tan⁡2x}=f(x)−1\diff{}{x}\{\tan^2 x\} = f(x)-1

Who ended up with the correct answer?

Hint

Recall tan⁡2x+1=sec⁡2x\tan^2x+1=\sec^2 x.

Answer

Both students.

Full solution

Recall that “+C+C" means that we can add any constant to the function. Since tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1, Students A and B have equivalent answers: they only differ by a constant.

So, if one is right, both are right; if one is wrong, both are wrong. We check Student A's work:

ddx{tan⁡2x+x+C}=ddx{tan⁡2x}+1+0=f(x)−1+1=f(x)\diff{}{x}\{\tan^2 x + x + C\}=\diff{}{x}\{\tan^2 x\} +1 + 0 = f(x)-1+1=f(x)

So, Student A's answer is indeed an anditerivative of f(x)f(x). Therefore, both students ended up with the correct answer.

Remark: it is a frequent occurrence that equivalent answers might look quite different. As you are comparing your work to others', this is a good thing to keep in mind!

Q51Stage 3

Let F(x)=∫0xx3sin⁡(t) dtF(x)=\displaystyle\int_0^x x^3 \sin(t)\ \dee{t}.

  1. Evaluate F(3)F(3).

  2. What is F′(x)F'(x)?

Hint

Since the integration is with respect to tt, the x3x^3 term can be moved outside the integral.

Answer

(a) 27(1−cos⁡3)27(1-\cos 3 ) (b) x3sin⁡(x)+3x2[1−cos⁡(x)]x^3\sin (x) + 3x^2[1-\cos (x)]

Full solution

  1. When x=3x=3,

    F(3)=∫0333sin⁡(t) dt=27∫03sin⁡t dt\begin{align*}F(3)&=\displaystyle\int_0^3 3^3 \sin(t)\ \dee{t}=27\int_0^3 \sin t \ \dee{t}\end{align*}

    Using the Fundamental Theorem of Calculus Part 2,

    =27[−cos⁡t]t=0t=3=27[−cos⁡3−(−cos⁡0)]=27(1−cos⁡3)\begin{align*}&=27\left[-\cos t\right]_{t=0}^{t=3} =27\left[-\cos 3 - (-\cos 0)\right]\\ &=27(1-\cos 3 )\end{align*}

  2. Since the integration is with respect to tt, the x3x^3 term can be moved outside the integral. That is: for the purposes of the integral, x3x^3 is a constant (although for the purposes of the derivative, it certainly is not).

    F(x)=∫0xx3sin⁡(t) dt=x3∫0xsin⁡(t) dt\begin{align*}F(x)&=\displaystyle\int_0^x x^3 \sin(t)\ \dee{t} = x^3 \int_0^x \sin(t)\ \dee{t}\end{align*}

    Using the product rule and the Fundamental Theorem of Calculus Part 1,

    F′(x)=x3⋅sin⁡(x)+3x2∫0xsin⁡(t) dt=x3sin⁡(x)+3x2[−cos⁡(t)]t=0t=x=x3sin⁡(x)+3x2[−cos⁡(x)−(−cos⁡(0))]=x3sin⁡(x)+3x2[1−cos⁡(x)]\begin{align*}F'(x)&=x^3\cdot \sin(x) + 3x^2 \int_0^x \sin(t)\ \dee{t}\\ &=x^3\sin(x)+3x^2\left[-\cos(t)\right]_{t=0}^{t=x}\\ &=x^3\sin(x)+3x^2[-\cos(x)-(-\cos(0))]\\ &=x^3\sin (x) + 3x^2[1-\cos (x)]\end{align*}

Remark: Since xx and tt play different roles in our problem, it's crucial that they have different names. This is one reason why we should avoid the common mistake of writing ∫axf(x)dx\int_a^x f(x)\dee{x} when we mean ∫axf(t)dt\int_a^x f(t)\dee{t}.

Q52Stage 3

Let f(x)f(x) be an even function, defined everywhere, and let F(x)F(x) be an antiderivative of f(x)f(x). Is F(x)F(x) even, odd, or not necessarily either one? (You may use your answer from Section 1.2, Question 20. )

Hint

Remember that antiderivatives may have a constant term.

Answer

If f(x)=0f(x)=0 for all xx, then F(x)F(x) is even and possibly also odd.

If f(x)≠0f(x) \neq 0 for some xx, then F(x)F(x) is not even. It might be odd, and it might be neither even nor odd.

(Perhaps surprisingly, every antiderivative of an odd function is even.)

Full solution

If F(x)F(x) is even, then f(x)f(x) is odd (by the result of Question 20 in Section 1.2). So, F(x)F(x) can only be even if f(x)f(x) is both even and odd. By the result in Question 19, Section 1.2, this means F(x)F(x) is only even if f(x)=0f(x)=0 for all xx. Note if f(x)=0f(x)=0, then F(x)F(x) is a constant function. So, it is certainly even, and it might be odd as well if F(x)=f(x)=0F(x)=f(x)=0.

Therefore, if f(x)≠0f(x) \neq 0 for some xx, then F(x)F(x) is not even. It could be odd, or it could be neither even nor odd. We can come up with examples of both types: if f(x)=1f(x)=1, then F(x)=xF(x)=x is an odd antiderivative, and F(x)=x+1F(x)=x+1 is an antiderivative that is neither even nor odd.

Interestingly, the antiderivative of an odd function is always even. The proof is a little beyond what we might ask you, but is given below for completeness. The proof goes like this: First, we'll show that if g(x)g(x) is odd, then there is some antiderivative of g(x)g(x) that is even. Then, we'll show that every antiderivative of g(x)g(x) is even.

So, suppose g(x)g(x) is odd and define G(x)=∫0xg(t)dtG(x)=\displaystyle\int_0^x g(t)\dee{t}. By the Fundamental Theorem of Calculus Part 1, G′(x)=g(x)G'(x)=g(x), so G(x)G(x) is an antiderivative of g(x)g(x). Since g(x)g(x) is odd, for any x≥0x\ge 0, the net signed area under the curve along [0,x][0,x] is the negative of the net signed area under the curve along [−x,0][-x,0]. So,

∫0xg(t) dt=−∫−x0g(t) dt(See Example 1.2.10 in the CLP-2 text)=∫0−xg(t) dt\begin{alignat*}{2}\int_0^x g(t)\ \dee{}t &= - \int_{-x}^0 g(t)\ \dee{t}&\quad\text{(See Example~\text{1.2.10} in the CLP-2 text)}\\ &=\int_0^{-x} g(t)\ \dee{t}\end{alignat*}

By the definition of G(x)G(x),

G(x)=G(−x)\begin{alignat*}{2}G(x)&=G(-x)\end{alignat*}

That is, G(x)G(x) is even. We've shown that there exists some antiderivative of g(x)g(x) that is even; it remains to show that all of them are even.

Recall that every antiderivative of g(x)g(x) differs from G(x)G(x) by some constant. So, any antiderivative of g(x)g(x) can be written as G(x)+CG(x)+C, and G(−x)+C=G(x)+CG(-x)+C = G(x)+C. So, every antiderivative of an odd function is even.

My list

nothing marked yet

Loading…

Open the whole list →

Your tutor can open this list with you. It follows your account, so it is there on whichever device you study on.

From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.