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.

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

500 km

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

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

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

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

CLP-2 text.

$1$

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

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

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

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

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 $k$.

$\dfrac{2}{\pi}$

The method of partial fractions can help you antidifferentiate.

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

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

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

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

$\dfrac12$

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%

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) $\sqrt3$

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

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

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) $65\text{ km/hr}$

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) $\displaystyle4-2e+2(e-1)\log(e-1) \approx 0.42$

This is an application of the ideas in Question 20.

(a) neither–both are zero (b) $|f(x)-A|$ has the larger average on $[0,4]$

Slice the solid into circular disks of radius $|f(x)|$ and thickness $\dee{x}$.

$(b-a)\pi R^2$

The question tells you $\frac{1}{1-0}\int_0^1 f(x)\,\dee{x} = \frac{f(0)+f(1)}{2}$.

0

Set up this question just like Question 23, but with variables for your limits of integration.

Note $(s-t)^2=s^2-2st+t^2$.

Yes, but if $a\neq0$, then $s=t$.

What are the graphs of $f(x)$ and $f(a+b-x)$ like?

$A$

For (b), express $A(x)$ as an integral, then differentiate.

(a) $\displaystyle\frac{bA(b) - aA(a)}{b-a}$ (b) $f(t)=A(t)+tA'(t)$

For (b), consider the cases that $f(x)$ is always bigger or always smaller than 0. Then, use the intermediate value theorem, Theorem 1.6.12 in the CLP-1 text.

Try l'H^opital's rule.

true

Use the result of Question 28.

0