1.3 The Fundamental Theorem of Calculus

$e^2-e^{-2}$

First find the general antiderivative by guessing and checking.

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

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

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

Check by differentiating.

false

Check by differentiating.

false

Use the Fundamental Theorem of Calculus Part 1.

$\sin(x^2)$

Use the Fundamental Theorem of Calculus, Part 1.

$\sqrt[3]{e}$

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

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

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

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

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

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

$F(x)$ represents net signed area.

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

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

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

Using the definition of the derivative, $F'(x) = \displaystyle\lim_{h \to 0}\dfrac{F(x+h)-F(x)}{h}$.

The area of a trapezoid with base $b$ and heights $h_1$ and $h_2$ is $\frac{1}{2}b(h_1+h_2)$.

Using the definition of the derivative,

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

So, $F'(x)=x$.

There is only one!

$f(t)=0$

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

$\int \log(ax)\ \dee{x}= x\log(ax)-x+C$, where $a$ is a given constant, and $C$ is any constant.

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

$\int x^3e^x\ \dee{x}=e^x\left(x^3-3x^2+6x-6\right)+C$

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

$\displaystyle\int \dfrac{1}{\sqrt{x^2+a^2}}\ \dee{x} = \log\left|x+\sqrt{x^2+a^2}\right|+C$ when $a$ is a given constant. As usual, $C$ is an arbitrary constant.

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

$\displaystyle\int \dfrac{x}{\sqrt{x(a+x)}}\ \dee{x}=\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)+C$

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

$5-\cos 2$

Split the given integral up into two integrals.

$2$

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

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

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

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

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

$\tan x - x +C$

Recall $2\sin x \cos x = \sin(2x)$.

$-\dfrac{3}{4}\cos(2x)+C$, or equivalently, $\dfrac{3}{2}\sin^2 x+C$

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

$\dfrac{1}{2}x+\dfrac{1}{4}\sin(2x)+C$

$F'\left(\frac{\pi}{2}\right)=\log(3)$ $G'\left(\frac{\pi}{2}\right)=-\log(3)$

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.

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

See Example 1.3.5 in the

CLP-2 text.

$F'(x)=-\dfrac{\sin x}{\cos^3x+6}$

See Example 1.3.5 in the

CLP-2 text.

$4x^3e^{(1+x^4)^2}$

See Example 1.3.5 in the

CLP-2 text.

$\big(\sin^6 x+8)\cos x$

See Example 1.3.5 in the

CLP-2 text.

$F'(1)=3e^{-1}$

See Example 1.3.6 in the

CLP-2 text.

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

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

$f(x)=2x$

What is the title of this section?

$f(4)=4\pi$

See Example 1.3.6 in the

CLP-2 text.

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

See Example 1.3.6 in the CLP-2 text.

$e^{\sin x}-e^{\sin(x^4-x^3)}\big(4x^3-3x^2\big)$

See Example 1.3.6 in the CLP-2 text.

$-2x \cos\big(e^{-x^2}\big) -5x^4\cos\big(e^{x^5}\big)$

See Example 1.3.6 in the CLP-2 text.

$e^x\sqrt{\sin(e^x)} -\sqrt{\sin(x)}$

Split up the domain of integration.

$14$

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

$\dfrac{5}{2}$

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

$45\,\textrm{m}$

See Example 1.3.5 in the

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

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

See Example 1.3.5 in the

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

The minimum is $\int_0^{-1} \frac{\dee{t}}{1+t^4}$. As $x$ runs from $-\infty$ to $\infty$, the function $f(x)= \int_0^{x^2-2x}\frac{\dee{t}}{1+t^4}$ decreases until $x$ reaches 1 and then increases all $x>1$. So the minimum is achieved for $x=1$. At $x=1$, $x^2-2x=-1$.

See Example 1.3.5 in the

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

$F$ achieves its maximum value at $x=\pi$.

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

CLP-2 text.

$2$

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

$\log 2$

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

Is $F(x)$ even or odd?

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

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

(a) $3x^2 \displaystyle\int_{0}^{x^3+1} e^{t^3} \dee{t} + 3x^5 e^{(x^3+1)^3}$ (b) $y = -3(x+1)$

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

Both students.

Since the integration is with respect to $t$, the $x^3$ term can be moved outside the integral.

(a) $27(1-\cos 3 )$ (b) $x^3\sin (x) + 3x^2[1-\cos (x)]$

Remember that antiderivatives may have a constant term.

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

If $f(x) \neq 0$ for some $x$, then $F(x)$ 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.)