Test 2 review (updated, now with answers and some hints)

Test 2 is scheduled for the first hour or so of class on Wednesday 25 October.

Test 2 review self-tests here: MAT1575Test2Review 

You probably should allow 90 minutes per self-test: the test itself will not have quite so many problems in it!

Important note: there are some errors in a couple of the problems, which I did not catch before publishing these:

In Self-Test 1, problem 4, the integral should go from 7 to 9 (not 4 to 9)

In Self-Test 2, problem 2, there is a typesetting error: the integral should read \displaystyle \int \frac{x}{\sqrt{9x^{2}-1}}\textrm{d}x

Answers and hints are at the end of this post, after the “more”

Topics:
• Trigonometric integrals of the form $\int \tan^{m}(x) \sec^{n}(x)\textrm{d}x$, where m and n are positive integers.
• Trigonometric substitution, including completing the square to use it
• Partial Fractions, including repeated factors and irreducible quadratic factors
• Improper integrals and the two comparison theorems

Also make sure that you are familiar with the additional antiderivatives we found and added to our basic list:

MAT1575Integration-Formulas-Theorems-OurMethods-slideshow

 

It may be useful to fill out the Outline Summary of Integration Methods:

MAT1575OutlineSummaryOfIntegrationMethods

 

Here is a nice resource: an online, free Integral calculator. Don’t let it do everything for you, but it’s great for checking your work. This can be very helpful in locating errors since it shows step-by-step how it arrived at the integrals.

Answers and hints: Please let me know if you find any typos in these! 

 

Self-Test 1

1) $latex  \frac{1}{3}\tan^{3}(x) – \tan(x) + x +C$

2) \frac{1}{25}\ln|x+2| - \frac{1}{25}\ln|x-3| -\frac{1}{5(x-3)} + C

3) Use trig substitution x = 3\sin\theta: the integral is

\frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) + \frac{1}{2}x\sqrt{9-x^{2}} + C

4) (Remember to change the bounds of integration so that you integrate from 7 to 9)

It’s possible to find this integral by completing the square and using trig substitution, or else you can factor the denominator (not easy! you must use the quadratic formula) and use partial fractions.

The integral is -\frac{1}{\sqrt{7}}\left(\ln\left(\frac{\left(5+\sqrt{7}\right)}{3\sqrt{2}}\right)-\ln\left(\frac{\left(3+\sqrt{7}\right)}{\sqrt{2}}\right)\right)

5) \sqrt{2}-\frac{\ln(\sqrt{2}+1)}{2}+\frac{\ln(\sqrt{2}-1)}{2}

6) The integral converges to \frac{1}{2e}

7) The integral diverges: you can use direct comparison, \frac{x+3}{x^{2}} > \frac{1}{x} and \displaystyle \int_{1}^{\infty}\frac{1}{x}\textrm{d}x diverges.

8) \displaystyle \int_{0}^{2}\frac{1}{(x-1)^{2}}\textrm{d}x = \lim_{c\rightarrow 1^{-}}\int_{0}^{c}\frac{1}{(x-1)^{2}}\textrm{d}x + \lim_{c\rightarrow 1^{+}}\int_{c}^{2}\frac{1}{(x-1)^{2}}\textrm{d}x

and those two limits diverge (you can find the antiderivative and show this directly), so the integral diverges.

 

Self-test 2

1) \frac{1}{4}\sec^{4}(x) + C

2) \frac{\sqrt{9x^{2}-1}}{9} + C

3) \frac{1}{4}\ln|x-1| - \frac{1}{8}\ln\left(x^{2}+3\right) - \frac{1}{4\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + C

4) \frac{27}{4}

5) Use partial fractions, or else complete the square and use trig substitution: the integral is 2\ln(3) -\ln(4)-\ln(2).

6) The integral diverges: the antiderivative is \ln|x| that goes to infinity as x\rightarrow -\infty.

7) As with problem 8 in self-test 1, you have to break this up into two limits because of the singularity at \frac{\pi}{2}: the integrals diverge.

8) The integral converges by direct comparison, but we have to work to get there because the comparison tests only hold for functions which are non-negative: so notice that whatever \displaystyle \int_{2}^{\infty}\frac{\sin(x)}{x^{2}}\textrm{d}x is, if it converges, its value will be less than \displaystyle \int_{2}^{\infty}\frac{\sin(x)|}{x^{2}}\textrm{d}x.

 

0\le\frac{\sin(x)|}{x^{2}} \le \frac{1}{x^{2}} for x>2, and the integral of \frac{1}{x^{2}} converges, so that means that \displaystyle \int_{2}^{\infty}\frac{\sin(x)|}{x^{2}}\textrm{d}x converges, and since the integral \displaystyle \int_{2}^{\infty}\frac{\sin(x)}{x^{2}}\textrm{d}x has to be less than that value, it must also converge.