2.2 Averages

See Definition 2.2.2 in the CLP-2 text and the discussion following it for the link between area under the curve and averages.

The most straightforward of many possible answers is shown.

Since the average of $f(x)$ on the interval $[0,5]$ is $A$, using Definition 2.2.2 in the CLP-2 text,

So, a rectangle with width 5 and height $A$ has area $\int_0^5 f(x)\,\dee{x}$.

That is: if we replace $f(x)$ with the constant function $g(x)=A$, then on the interval $[0,5]$, the area under the curve is unchanged.

(There are many rectangles with area $5A$; we drew the one we consider to be the most straightforward in this context.)

Average velocity is discussed in Example 2.2.5 of the CLP-2 text. You don't need an integral for this.

500 km

Average velocity, as discussed in Example 2.2.5 of the CLP-2 text, is change in position divided by change in time. So, the change in position (i.e. distance travelled) is (100 km/h)(5 h) = 500 km.

Much like Problem 2, you don't need to do any integration here.

$\dfrac{W}{b-a}$ N

The work done is

so the average value of $F(x)$ is

We can quickly check our units: since $W$ is in joules (that is, newton-metres), and $b-a$ is in metres, so $\frac{W}{b-a}$ is in newtons.

Part (a) is asking the length of the pieces we've cut our interval into. Part (c) should be given in terms of $f$. Our final answer in (d) will resemble a Riemann sum, but without some extra manipulation it won't be in exactly the form of a Riemann sum we're used to.

(a) $\dfrac{b-a}{n}$ (b) $a+3\dfrac{b-a}{n}$ (c) $f\left(a+3\dfrac{b-a}{n} \right)$ (d) $\dfrac{1}{n}\sum\limits_{i=1}^n f\left(a+(i-1)\frac{b-a}{n}\right)$

For (b), the value of $f(0)$ could be much, much larger than $g(0)$.

(a) yes (b) not enough information

The answer is something very simple.

0

Recall the definition of an odd function: $f(-x)=-f(x)$. Since the domain of integration is symmetric, the signed area on one side of the $y$-axis “cancels out" the signed area on the other–this is Theorem 1.2.11 in the CLP-2 text.

Apply the definition of “average value” in Section 2.2 of the

CLP-2 text.

$1$

By definition, the average value is

We now observe that $\sin(5x)$ is an odd function, and hence its integral over the symmetric interval $[-\frac\pi2,\frac\pi2]$ equals zero. So the average value of $f(x)$ on this interval is $1$:

Alternatively, using the fundamental theorem of calculus, the average equals:

You can antidifferentiate $x^2\log x$ using integration by parts.

$\displaystyle\frac{1}{e-1}\Big[\frac{2}{9}e^3+\frac{1}{9}\Big]$

By definition, the average is

To antidifferentiate, we use integration by parts with $u=\log x$ and $\dee{v}=x^2\,\dee{x}$, hence $\dee{u}=\frac{1}{x}\,\dee{x}$ and $v=\frac{1}{3}x^3$.

You can antidifferentiate an odd power of cosine with a substitution; for an even power of cosine, use the identity $\cos^2 x = \frac{1}{2}\big(1+\cos(2x)\big)$.

$\dfrac{4}{\pi}+1$

By definition, the average value in question equals

For the first integral we use the substitution $u = \sin x$, $\dee{u}=\cos x\,\dee{x}$, $\cos^2 x=1-\sin^2 x = 1-u^2$. Note that the endpoints $x=0$ and $x=\frac\pi2$ become $u=0$ and $u=1$, respectively.

For the second integral we use the trigonometric identity $\cos^2x\,\dee{x} = \frac{1 + \cos(2x)}{2}$.

Therefore, the average value in question is

If you're not sure how to antidifferentiate, try the substitution $u=kx$, $\dee{u}=k\dee{x}$, keeping in mind that $k$ is a constant. Interestingly, your final answer won't depend on $k$.

$\dfrac{2}{\pi}$

By definition, the average value in question equals

To evaluate the integral, we use the substitution $u = kx$, $\dee{u}=k\,\dee{x}$. Note that the endpoints $x=0$ and $x=\pi/k$ become $u=0$ and $u=\pi$, respectively. So

Remark: the average does not depend on $k$. To see why this is, note that $\sin(kx)$ runs between $-1$ and $1$ as $x$ changes. When $x=0$, $kx=0$, and when $x=\pi/k$, $kx=\pi$. So, our function $\sin (kx)$ runs exactly from $\sin 0 =0$ to $\sin(\pi/2)=1$, then back down to $\sin\pi=0$.

The method of partial fractions can help you antidifferentiate.

$\dfrac{10}{3}\log 7$ degrees Celsius

By definition, the average temperature is

We don't see an obvious substitution, but integrand is a rational function. The degree of the numerator is strictly less than the degree of the denominator, so we factor the denominator and use a partial fraction decomposition.

Setting $x=4$, we see $80=8A$, so $A=10$. Setting $x=-4$, we see $80=8B$, so $B=10$.

Try the substitution $u=\log x$, $\dee{u}=\frac{1}{x}\,\dee{x}$.

$\dfrac{1}{2(e-1)}$

By definition, the average value is

To integrate, we use the substitution $u=\log x$, $\dee{u}=\frac{1}{x}\,\dee{x}$. Then the limits of integration become 0 and 1, respectively.

Remember $\cos^2 x = \frac{1}{2}\big(1+\cos(2x)\big)$.

$\dfrac12$

By definition, the average value is:

Notice the term $50\cos\left(\frac{t}{12}\pi\right)$ has a period of 24 hours, while the term $200\cos\left(\frac{t}{4380}\pi\right)$ has a period of one year.

If $n$ is an approximation of $c$, then the relative error of $n$ is $\frac{|n-c|}{c}$.

(a) 400 ppm (b) $\approx 599.99$ ppm (c) 0.125, or 12.5%

Before we start answering questions, let's look at our function a little more carefully. The term $50\cos\left(\frac{t}{12}\pi\right)$ has a period of 24 hours, while the term $200\cos\left(\frac{t}{4380}\pi\right)$ has a period of one year. So, the former term describes a standard daily variation, while the latter gives a seasonal variation over the year.

(a) Using the definition of an average, the average concentration over one year ($t=0$ to $8760$) is:

Since $\frac{8760}{12} = 730$, which is even, $\sin\left(\frac{8760}{12}\pi\right)=\sin(0)=0$. Also, $\sin\left(\frac{8760}{4380}\pi\right)=\sin(2\pi)=0$.

Remark: for the portions of the integral in red and blue, we also could have noticed that the integrand goes through a whole (integer) number of periods. For every period, the net signed area between the curve and the $x$-axis is zero, so we could have seen from the very beginning these terms would contribute 0 to the final average.

(b) Using the definition of an average, the average concentration over the first day ($t=0$ to $t=24$) is:

Note $t=0$ to $t=24$ is one complete period for the integrand in red, so the red integral will evaluate to zero. However, $t=0$ to $t=24$ is less than one cycle for the integrand in blue, so we expect this will contribute some non-zero quantity to the average.

Remark: $C(0) = 400+\textcolor{red}{50}+\textcolor{blue}{200}$. The red term comes from the daily variation, and over the first day this will have an average of 0. The blue term comes from the seasonal variation, which changes dramatically over the course of an entire year but won't change very much over the course of a single day. So, it is reasonable that the average concentration over the first day should be close to (but not exactly) $400+200$ ppm.

(c) The average of $N(t)$ over $[0,8760]$ is:

Since the average of $C(t)$ was 400, this gives us an absolute error of $|400-350|=50$ ppm, for a relative error of

or 12.5%.

That is: sampling at the same time every day, rather than throughout the day, lead to an error of 12.5% in the yearly average concentration of carbon dioxide.

A cross section of $S$ at location $x$ is a circle with radius $x^2$, so area $\pi x^4$. Part (a) is asking for the average of this function on $[0,2]$.

(a) $\dfrac{16\pi}{5}$ (b) $\dfrac{32\pi}{5}$ (c) $\dfrac{32\pi}{5}$

(a) can be done without calculation

(a) 0 (b) $\sqrt3$

$\tan^2 x = \sec^2 x - 1$

$\displaystyle\sqrt{\frac{4}{\pi} - 1}\approx 0.52$

Using the definition provided,

Remember force is the product of the spring constant with the distance it's stretched past its natural length. The units given in the question are not exactly standard, but they are compatible with each other.

You can find part (b) without any calculation. For (c), remember $\sin^2 x = \frac{1}{2}\big(1-\cos(2x)\big)$.

(a) $F(t) = 3f(t) = 3\sin\left(t\pi\right)$ N (b) 0 (c) $\dfrac{3}{\sqrt{2}} \approx 2.12$

The trapezoidal rule is found in Section 1.11.2 of the

CLP-2 text.

(a) $130\text{ km}$ (b) $65\text{ km/hr}$

(a) Let $v(t)$ be the speed of the car at time $t$. Then, by the trapezoidal rule with $a=0$, $b=2$, $\De t=1/3$, the distance traveled is

(b) The average speed is $\frac{\text{dist}}{\text{time}} \approx \frac{130\text{ km}}{2\text{ hr}} = 65\text{ km/hr}$.

To find a definite integral of the absolute value of a function, break up the interval of integration into regions where the function is positive, and intervals where it's negative.

(a) $A=e-1$ (b) 0 (c) $\displaystyle4-2e+2(e-1)\log(e-1) \approx 0.42$