Exercises: Cumulative

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Exercises relating to various topics we have studied.

A key skill to practice is deciding which of the several techniques we have learned should be applied to a particular problem. In these exercises and questions, you will be responsible for choosing a method which works for you.

Compute the indefinite integral. $\int \ln x \, dx = \answer { x \ln x - x} + C$ (You should choose an answer which equals $-1$ at $x = 1$ ; if your answer does not satisfy this, add a constant to make it so.)

Integrate by parts.

Let $u = \ln x$ and $dv = 1$ .

Compute the indefinite integral. $\int \frac {dx}{(x^2+1)^2} = \answer { \frac {1}{2} \frac {x}{x^2+1} + \frac {1}{2} \arctan x} + C.$ (You should choose an answer which equals $0$ at $x = 0$ ; if your answer does not satisfy this, add a constant to make it so.)

Partial fractions won’t help in this case because the expression is already simplified as much as possible.

Try a trigonometric substitution

Make the substitution $x = \tan \theta$ .

You should arrive at the integral $\int \cos ^2 \theta \, d \theta$ after the substitution; here you’ll need to use a power-reduction formula to continue.

Sample Quiz Questions

The region in the plane between the $x$ -axis and the graph $y = \frac {1}{2} \ln {x} + 1$ in the range $\frac {1}{5} \leq x \leq 1$ is revolved around the axis $x = \frac {1}{10}$ . Compute the volume of the resulting solid.

$\displaystyle \frac {11}{25} \pi$ $\displaystyle \frac {14}{25} \pi$ $\displaystyle \frac {16}{25} \pi$ $\displaystyle \frac {19}{25} \pi$ $\displaystyle \frac {21}{25} \pi$ $\displaystyle \frac {24}{25} \pi$

If the variable $x$ is used for slicing, then slices are parallel to the axis of rotation, which indicates the shell method should be used. The radius of a shell is $x - \frac {1}{10}$ . The height of a shell is exactly $\frac {1}{2} \ln {x} + 1$ . The volume of the region is therefore given by $\int _{\frac {1}{5}}^{1} \frac {\pi }{10} \left (10 x - 1\right ) \left (\ln {x} + 2\right )\, dx.$ To compute the integral, we can use integration by parts. A reasonable strategy is to integrate $x - \frac {1}{10}$ and differentiate $\ln {x} + 2$ . This gives the equality $\begin {aligned} \pi \int \left (x - \frac {1}{10}\right ) \left (\ln {x} + 2\right )\, dx & = \pi \left (\left (\frac {x^{2}}{2} - \frac {x}{10}\right ) \left (\ln {x} + 2\right ) - \int \frac {1}{x} \left (\frac {x^{2}}{2} - \frac {x}{10}\right )\, dx\right ) \\ & = - \pi \left (\frac {x^{2}}{4} - \frac {x}{10}\right ) + \pi \left (\frac {x^{2}}{2} - \frac {x}{10}\right ) \left (\ln {x} + 2\right ). \end {aligned}$ Therefore $\begin {aligned} \pi \int _{\frac {1}{5}}^{1} \left (x - \frac {1}{10}\right ) \left (\ln {x} + 2\right )\, dx & = \left . \left [- \pi \left (\frac {x^{2}}{4} - \frac {x}{10}\right ) + \pi \left (\frac {x^{2}}{2} - \frac {x}{10}\right ) \left (\ln {x} + 2\right ) \right ] \right |_{\frac {1}{5}}^{1}\\ & = \left (\frac {13}{20} \pi \right ) - \left (\frac {\pi }{100} \right ) = \frac {16}{25} \pi . \end {aligned}$

Consider the region given by $2 \pi \leq x \leq \frac {5}{2} \pi$ and $0 \leq y \leq \sin {x}$ . Compute the $x$ -coordinate of the centroid (i.e., assuming constant density).

$\displaystyle -1 + \frac {5}{2} \pi$ $\displaystyle 1 + 2 \pi$ $\displaystyle \frac {5}{2} \pi$ $\displaystyle -1 + 3 \pi$ $\displaystyle 3 \pi$ $\displaystyle 4 \pi$

The mass $M$ will be given by the integral $\int _{2 \pi }^{\frac {5}{2} \pi } \sin {x}\, dx$ One can check that $\begin {aligned} \int _{2 \pi }^{\frac {5}{2} \pi } \sin {x}\, dx & = 1. \end {aligned}$ To compute the $x$ -coordinate of the centroid, we also need to compute the integral $\int _{2 \pi }^{\frac {5}{2} \pi } x \sin {x}\, dx$ To compute the integral, we can use integration by parts. A reasonable strategy is to integrate $\sin {x}$ and differentiate $x$ . This gives the equality $\begin {aligned} \int x \sin {x}\, dx & = - x \cos {x} - \int \left (- \cos {x}\right )\, dx \\ & = - x \cos {x} + \sin {x}. \end {aligned}$ Therefore $\begin {aligned} \int _{2 \pi }^{\frac {5}{2} \pi } x \sin {x}\, dx & = \left . \left [- x \cos {x} + \sin {x} \right ] \right |_{2 \pi }^{\frac {5}{2} \pi }\\ & = 1 - \left (- 2 \pi \right ) = 1 + 2 \pi . \end {aligned}$ The corret answer is the ratio of the integrals, i.e., $\begin {aligned} \overline {x} & = \frac {1 + 2 \pi }{1} = 1 + 2 \pi . \end {aligned}$

Sample Exam Questions

An object moves in such a way that its acceleration at time $t$ seconds is $(t^2 + 5t + 6)^{-1}$ meters per second squared. If the initial velocity of the object is $2/3$ meters per second, what is the limit of its velocity as $t \rightarrow \infty$ ?

$\displaystyle \ln \frac {3}{2}$ meters per second $\displaystyle \ln 6$ meters per second $1$ meters per second $\displaystyle \ln \frac {4}{9}$ meters per second $\displaystyle \ln \frac {9}{4}$ meters per second $0$ meters per second

Find the volume of the solid generated by revolving the region bounded above by $y = \sin x$ and bounded below by $y = 0$ for $0 \leq x \leq \pi$ about the line $x = \pi$ .

$\pi ^2$ $2\pi ^2$ $4\pi ^2$ $\displaystyle \frac {\pi ^2}{2}$ $\displaystyle \frac {\pi ^2}{4}$ none of these

Evaluate $\displaystyle \int _1^2 x \ln (x^2 + 1) dx$ .

$0$ $1$ $\ln 2$ $\displaystyle \frac {1}{2}$ $\displaystyle \ln (2) - \frac {1}{2}$ none of these

This integral can be computed via integration by parts. If we integrate $x$ and differentiate $\ln (x^2+1)$ , we get $\begin {aligned} \int _1^2 x \ln (x^2+1) dx & = \left . \frac {x^2}{2} \ln (x^2+1) \right |_{1}^2 - \int _1^2 \frac {x^2}{2} \frac {2x}{x^2+1} dx \\ & = 2 \ln 5 - \frac {1}{2} \ln 2 - \int _1^2 \frac {x^3}{x^2+1} dx. \end {aligned}$ The latter integral can be simplified using polynomial long division: $\displaystyle \frac {x^3}{x^2+1} = x - \frac {x}{x^2+1}$ . Therefore $\begin {aligned} \int _1^2 x \ln (x^2+1) dx & = 2 \ln 5 - \frac {1}{2} \ln 2 - \int _1^2 x dx + \int _1^2 \frac {x}{x^2+1} dx \\ & = 2 \ln 5 - \frac {1}{2} \ln 2 - \left . \frac {x^2}{2} \right |_1^2 + \left . \frac {1}{2} \ln (x^2+1) \right |_1^2 \\ & = 2 \ln 5 - \frac {1}{2} \ln 2 - 2 + \frac {1}{2} + \frac {1}{2} \ln 5 - \frac {1}{2} \ln 2 \\ & = \frac {5}{2} \ln 5 - \frac {2}{2} \ln 4 - \frac {3}{2} - \frac {3}{2} = \ln \left ( \frac {5^5}{4} \right ) - \frac {3}{2}. \end {aligned}$