Practice Exam: Numerical Integration, Improper Integrals, Applications
Time: 60 minutes


Problem 1 (15 points)   Compute the exact value of $\displaystyle \int_0^\infty x e^{-2x^2}\,dx $

Solution: The integrand is continuous for all x, consequently the only "impropriety" occurs at $\infty$.

\begin{eqnarray*}\int_0^\infty x e^{-2x^2}\,dx&=&\lim_{N\to\infty}\int_0^N x e^{...
...m_{N\to\infty}-\frac{1}{4}\left(e^{-2N^2}-e^0\right)=\frac{1}{4}
\end{eqnarray*}


In the last step we use the fact that $\displaystyle \lim_{N\to\infty}e^{-2N^2}=0 $.

Problem 2 (15 points)   To approximate the integral $\displaystyle \int_a^b f(x)\,dx $, the interval [a,b] is divided into n parts of length $\Delta x$ each. Explain why

\begin{displaymath}\mbox{TRAP}(n)=\mbox{LEFT}(n)+\frac{1}{2}(f(b)-f(a))\cdot\Delta x\end{displaymath}

Solution: Let $\displaystyle \{a=x_0<x_1<x_2<\cdots<x_n=b\} $ denote the endpoints of the n subintervals. Then

\begin{displaymath}\mbox{LEFT}(n)=f(a) \Delta x+\sum_{k=1}^{n-1} f(x_k) \Delta x.\end{displaymath}

On the other hand,

\begin{displaymath}\mbox{RIGHT}(n)=\sum_{k=1}^{n-1} f(x_k) \Delta x +f(b) \Delta x.\end{displaymath}

Consequently,

\begin{displaymath}\mbox{RIGHT}(n)=\mbox{LEFT}(n)-f(a)\Delta x+f(b) \Delta x.\end{displaymath}

It follows that

\begin{eqnarray*}\mbox{TRAP}(n)&=&\frac{1}{2}\mbox{RIGHT}(n)+\frac{1}{2}\mbox{LE...
...ox{LEFT}(n)\\
&=&\mbox{LEFT}(n)+\frac{1}{2}(f(b)-f(a))\Delta x.
\end{eqnarray*}


Problem 3 (20 points)   Do the following integrals converge or diverge? For full credit, you must explain how you arrive at your answer.
1.
$\displaystyle \int_{0}^\infty \frac{x^2-4x+7}{(x^2+6)^2}\,dx $

Solution: The integrand is continuous for all $x\geq 0$, consequently the only "impropriety" occurs at $\infty$. For large x-values,

\begin{displaymath}\frac{x^2-4x+7}{(x^2+6)^2}\approx\frac{1}{x^2}.\end{displaymath}

(You may use the Limit Comparison Test for a more formal argument.) Since $\displaystyle \int_0^\infty \frac{dx}{x^2} $ converges at infinity by the p-test, so does the integral in question.
2.
$\displaystyle \int_3^\infty \frac{(x-2)\ln x}{x^2}\,dx $

Solution: The integrand is continuous for all $x\geq 3$, consequently the only "impropriety" occurs at $\infty$. For large x-values,

\begin{displaymath}\frac{(x-2)\ln x}{x^2}\approx\frac{x \ln x}{x^2}=\frac{\ln x}{x}.\end{displaymath}

(You may use the Limit Comparison Test for a more formal argument.)

For $x\geq 3$,

\begin{displaymath}\frac{\ln x}{x}\geq \frac{1}{x},\end{displaymath}

and thus by the Comparison Test, the integral in question is divergent, since $\displaystyle \int_3^\infty\frac{dx}{x} $ diverges.

Problem 4 (15 points)   For p>0 consider the improper integral

\begin{displaymath}\int_{0}^1 \frac{dx}{x^p}.\end{displaymath}

Derive the p-test for these integrals, i.e., find the values of p for which this integral converges, and the values for which it diverges. (To obtain full credit, you must show all steps necessary to obtain your answer.)

Solution: For $p\not=1$, we have

\begin{displaymath}\int_{0}^1 \frac{dx}{x^p}=\lim_{t\to 0^+}\int_{t}^1 \frac{dx}...
...left.\left(\frac{1}{1-p}x^{1-p}\right)\right\vert _{x=t}^{x=1}.\end{displaymath}

Since $\displaystyle \lim_{t\to 0^+}x^{1-p}=0 $ for p<1, while $\displaystyle \lim_{t\to 0^+}x^{1-p}=\infty $ for p>1, the integral converges for p<1 and diverges for p>1.

What if p=1? Then

\begin{displaymath}\int_{0}^1 \frac{dx}{x^p}=\lim_{t\to 0^+}\int_{t}^1 \frac{dx}{x}=-\lim_{t\to 0^+}\ln x=\infty,\end{displaymath}

so the integral diverges.

Problem 5 (15 points)   Let a,b>0. Find the volume of the ellipsoid you obtain, when you rotate the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $ about the y-axis.

Solution: Rotating the ellipse about the y-axis amounts to rotating the graph of $\displaystyle x=a \sqrt{1-\frac{y^2}{b^2}} $ about the y-axis. Each horizontal slice is a disk with area $\displaystyle \pi x^2=\pi a^2\left(1-\frac{y^2}{b^2}\right) $.

Consequently the volume of the ellipsoid is given by

\begin{displaymath}\int_{-b}^{b}\pi a^2 \left(1-\frac{y^2}{b^2}\right)\,dy=\frac{4}{3}\pi a^2 b.\end{displaymath}

Problem 6 (20 points)   A rectangular water reservoir is 50 km long and 30 km wide. The depth of the reservoir at each point is one tenth its distance to the nearest shore line. Find the volume of water in the reservoir (in km3).

View from the Top.
Darker blue corresponds to deeper water.

Solution: The maximal distance from any point in the reservoir to the nearest shoreline is 15 km; thus the maximal depth of the reservoir is 1.5 km. The distance to the shoreline is constant on lines parallel to the shorelines. More precisely, the water is at least h km deep inside a rectangle of width 30-20 h km and length 50-20h km. This leads to the Riemann sum approximation of the volume as

\begin{displaymath}\sum_{h=0}^{h=1.5} (30-20h)(50-20h) \Delta h,\end{displaymath}

with corresponding integral

\begin{displaymath}\int_{0}^{1.5} (30-20h)(50-20h)\,dh=900 \mbox{ km}^3.\end{displaymath}

View from the Side (upside down).
Darker blue corresponds to deeper water.


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