MATH6037: Review of Integration Exercise Solutions

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from https://jpmccarthymaths.wordpress.com/2011/02/01/math6037-general-information/

Question 1

(a)

Now:

$\frac{d}{dx}x^2=2x$ and $\frac{d}{dx}5x=5$ ,

hence:

$\int_0^1 (2x+5)\,dx=\left[x^2+5x\right]_0^1 =(1^2+5(1))-(0^2+5(0))=6$

(b)

There is no obvious anti-derivative and there is no obvious manipulation hence we are looking for the function-dervative pattern to make a substitution (or else use the LIATE rule). Notice that the top is the derivative of the bottom. Hence we will let $u$ be the function; i.e. the bottom:

Let $u=x^2+5x+1$ :

$\frac{du}{dx}=2x+5$

$\Rightarrow dx=\frac{du}{2x+5}$

Now put everything back into the integrand, suppressing the limits:

$I=\int \frac{2x+5}{u}\,\frac{du}{2x+5}$

$=\int \frac{du}{u}=\ln u =[\ln (x^2+5x+1)]_0^1$

$=\ln(1^2+5(1)+1)-\ln(0^2+5(0)+1)$

Now, what do we have to raise $e$ to, to get $1$ ? i.e. $e^?=1$ ; well $e^0=1$ . Hence $\ln 1=0$ :

$I=\ln 7$ .

(c)

Again no obvious anti-derivative or manipulation hence we are looking at making a substitution.

There are a number of options here. One is the straight substitution $u=2x+5$ . This will work out because $u'=2$ , just a constant. So a function, $2x+5$ and a multiple of it’s derivative are in the integrand. Alternatively the LIATE rule says pick $u=2x+5$ . Finally there are a number of other options once we know about terms of the form $a^{m+n}$ . We will pursue this here.

Now $a^{m}a^n=a^{m+n}$ , hence:

$I=\int_0^1e^{2x+5}\,dx=\int_0^1 e^{2x}.e^5\,dx$

Now $e^5$ is just a real number or scalar and because integration is essentially addition, we can take it out as a common term:

$I=e^5\int_0^1e^{2x}\,dx$

Now

$\frac{d}{dx}\frac{e^{ax}}{a}=\frac{ae^{ax}}{a}=e^{ax}$ (*)

hence

$I=e^5\left[\frac{e^{2x}}{2}\right]_0^1=e^5\left[\frac{e^2}{2}-\frac{e^0}{1}\right]$

Again $a^0=1$ ;

$I=\frac{e^5}{2}(e^2-1)$

Question 2

(a)

According to the hint, we should choose $u=1-x^2$ :

Let $u=1-x^2$ :

$\frac{du}{dx}=-2x$

$\Rightarrow dx=\frac{du}{-2x}$

Throwing this in, suppressing the limits for now:

$I=\int xu^5\frac{du}{-2x}=-\frac{1}{2}\int u^5\,du$

$I=-\frac{1}{2}\left[\frac{u^6}{6}\right]=-\frac{1}{2}\left[\frac{(1-x^2)^6}{6}\right]_0^1$

$I=-\frac{1}{2}\left[\frac{(1-1^2)^6}{6}-\frac{(1-0^2)^6}{6}\right]=\frac{1}{12}$

(b)

$I=\int_1^2\frac{dx}{e^x}=\int_1^2\ e^{-1x}\,dx$

Now using (*):

$I=\left[\frac{e^{-x}}{-1}\right]_1^2=(-e^{-2})-(-e^{-1})\approx 0.233$

using a calculator.

Question 3

First off apologies: the middle term should have been $\log 2$ .

This is a tough question but if you think carefully about your original definition of integration as finding the area under a curve, then we have that if $f(x)\leq g(x)$ then:

$\int_a^b f(x)\,dx\leq \int_a^b g(x)\,dx$

When $x\geq 1$ , $x^{-2}\leq x^{-1/2}\leq x^{-1}$ (in yellow, blue and red respectively). Hence the if we find the area under these curves between any two points bigger than 1, the area under a ‘smaller’ function will be smaller than the area under a ‘bigger’ one.

This is what we have to do here. The problem is between what limits should be pick. If divine inspiration doesn’t help I suggest solving the general problem; i.e. just let $a$ and $b$ be anything (well careful: we want to integrate where $x\geq 1$ , so we want $a,b\geq 1$ and also $a<b$ ). Now writing the three functions in their power form (i.e. $1/\sqrt{x}=1/x^{1/2}=x^{-1/2}$ ):

$\int_a^b x^{-2}\,dx\leq\int_a^b x^{-1}\,dx\leq\int_a^b x^{-1/2}\,dx$

$\Rightarrow \left[\frac{x^{-1}}{-1}\right]_a^b\leq\left[\ln x\right]_a^b\leq \left[\frac{x^{1/2}}{1/2}\right]_a^b$

$\Rightarrow \left[-b^{-1}-(-a^{-1})\right]\leq\left[\ln b-\ln a\right]\leq 2\left[\sqrt{b}-\sqrt{a}\right]$

To tidy up a little, using the log rule $\ln(a/b)=\ln a-\ln b$ :

$\frac{1}{a}-\frac{1}{b}\leq \ln (a/b)\leq 2(\sqrt{a}-\sqrt{b})$

This is valid for all $a,b\geq 1$ such that $a<b$ . How about $a=1,\,b=2$ ?

$\frac{1}{1}-\frac{1}{2}\leq\ln (2/1)\leq 2(\sqrt{2}-\sqrt{1})$

$\frac{1}{2}\leq \ln 2\leq 2\sqrt{2}-2$ $\bullet$

Question 4

(a)

Not in tables but try a manipulation:

$\frac{(x+1)^2}{2x}=\frac{(x+1)(x+1)}{2x}=\frac{x^2+x+x+1}{2x}=\frac{x^2+2x+1}{2x}$

If you are comfortable feel free to use the identity $(a+b)^2=a^2+2ab+b^2$ :

$\frac{x^2+2x+1}{2x}=\frac{x^2}{2x}+\frac{2x}{2x}+\frac{1}{2x}$

$=\frac{1}{2}x+1+\frac{1}{2}x^{-1}$ :

$I=\int_1^2 (\frac{1}{2}x+1+\frac{1}{2}x^{-1})\,dx$

$I=\frac{1}{2}\left[\frac{x^2}{2}\right]_1^2+[x]_1^2+\frac{1}{2}\left[\ln x\right]_1^2$

$I=\frac{1}{2}\left(\frac{2^2}{2}-\frac{1^2}{2}\right)+(2-1)+\frac{1}{2}(\ln 2-\ln 1)$

$I=\frac{1}{2}.\frac{3}{2}+1+\frac{1}{2}\ln 2=\frac{7}{4}+\ln \sqrt{2}$ .

(b)

No obvious anti-derivative or manipulation. Spot the function $x^2$ and derivative $(2)x$ .

Let $u=x^2$ :

$\frac{du}{dx}=2x$

$\Rightarrow dx=\frac{du}{2x}$

Again suppressing the limits

$\Rightarrow I=\int x\cos u\,\frac{du}{2x}=\frac{1}{2}\int \cos u\,du$

$I=\frac{1}{2}[\sin u]=\frac{1}{2}[\sin x^2]_0^{\sqrt{\pi/2}}$

$I=\frac{1}{2}\left[\sin (\pi/2)-\sin 0\right]$

Now using the calculator, tables, or preferably, the unit circle, $\sin (\pi/2)=1$ and $\sin 0=0$ :

$I=\frac{1}{2}$

(c)

There is no obvious anti-derivative (direct integration by rule) so we’re looking at a manipulation or a substitution. The trick here is to know that we know how to integrate:

$\int \frac{1}{\sqrt{a^2-x^2}}\,dx=\sin^{-1}\frac{x}{a}$ ,

and because $(4-x)(2+x)$ is of the form $a+bx-x^2$ , we can complete the square and write $(4-x)(2+x)=q^2-(x+p)^2$ and make the substitution $u=x+p$ .

$(4-x)(2+x)=8+4x-2x-x^2=8+2x-x^2$

Now find $p,q\in\mathbb{R}$ such that

$8+2x-x^2=q^2-(x+p)^2$

$=q^2-(x+p)(x+p)=q^2-(x^2+px+px+p^2)$

$=q^2-x^2-2px-p^2=(q^2-p^2)+(-2p)x-x^2$

Compare coefficients:

$8+2x-x^2=(q^2-p^2)+(-2p)x-x^2$

We need

$q^2-p^2=8$ and $-2p=2$

Hence $p=-1$ and $q^2=9=3^2$ . Now:

$I=\int_1^{5/2}\frac{1}{\sqrt{3^2-(x-1)^2}}\,dx$

Let $u=x-1$ :

$\frac{du}{dx}=1\Rightarrow du=dx$

Again suppressing the limits:

$I=\int\frac{1}{\sqrt{3^2-(u)^2}}\,du=\left[\sin^{-1}\frac{u}{3}\right]$

$I=\left[\sin^{-1}\frac{x-1}{3}\right]_1^{5/2}=\sin^{-1}(1/2)-\sin^{-1}0$

Now using a calculator, tables or preferably a little 30-60-90 right-angle-triangle diagram:

$I=\pi/6$

Question 5

Differentiation is linear. Let $k\in\mathbb{R}$ :

$\frac{d}{dx}(u(x)+kv(x))=u'(x)+kv'(x)$

Hence

$\frac{d}{dx}\frac{e^x+e^{-x}}{2}=\frac{1}{2}\left(\frac{d}{dx}e^{x}+\frac{d}{dx}e^{-x}\right)$

$f'(x)=\frac{1}{2}\left(e^x+e^{-x}(-1)\right)=g(x)$

There are two ways to finish this question. One is to note that, by the Chain Rule:

$\frac{d}{dx}[f(x)]^2=2f(x)f'(x)=2f(x)g(x),$

so $2f(x)g(x)$ has anti-derivative $[f(x)]^2$ . This is the obvious way.

Alternatively rewrite the integral:

$I=2\int_0^{\ln 1/2} f(x)f'(x)\,dx$ .

Now looking for the function-derivative pattern, let $u=f(x)$ :

$\frac{du}{dx}=f'(x)$

$\Rightarrow dx =\frac{du}{f'(x)}$

Suppressing the limits:

$I=\int u f'(x)\frac{du}{f'(x)}=2\int u\,du=[u^2/2]$

$=2[[f(x)]^2/2]_0^{\ln 1/2}$

The ‘2’s cancel. Putting in the limits:

$=\left[\left(\frac{e^{\ln 1/2}+e^{-\ln 1/2}}{2}\right)^2-\left(\frac{e^0+e^{-0}}{2}\right)^2\right]$

Now some facts about exponentials and logs:

$e^{\ln x}=x$ ,

$e^{-\ln x}=\frac{1}{e^{\ln x}}=\frac{1}{x}$ ,

$e^{0}=1$ .

Hence, with $1/(1/2)=2$ (multiply above and below by 2):

$I=\left(\frac{1/2+2}{2}\right)^2-\left(\frac{1+1}{2}\right)^2=\left(\frac{5}{4}\right)^2-1=\frac{9}{16}$ .

Question 6

(a)

We want to find $p,q\in\mathbb{R}$ such that

$x^2+4x+5=(x+p)^2+q^2$

$\Leftrightarrow x^2+4x+5=x^2+2px+p^2+q^2$

$\Leftrightarrow x^2+4x+5=x^2+2px+(p^2+q^2)$

$\Leftrightarrow 4=2p$ and $p^2+q^2=5$

$\Rightarrow p=2$ and hence $q=1$ .

So $x^2+4x+5=(x+2)^2+1^2$ . Now considering the integral

$\int\frac{1}{x^2+4x+5}\,dx=\int \frac{1}{(x+2)^2+1^1}$ ,

make the substitution $u=x+2$ :

$\frac{du}{dx}=1\Rightarrow dx=du$ :

$I=\int\frac{1}{u^2+1^2}\,du$

Now using the tables:

$I=\tan^{-1}u+c=\tan^{-1}(x+2)+C$

(b)

Apologies – this question requires partial fractions – which we didn’t do until Week 2.

Question 7

$1-\frac{e^x}{e^x+1}\frac{e^x}{(e^x+1)^2}$

$=1\times\underbrace{\frac{(e^x+1)^2}{(e^x+1)^2}}_{=1}-\frac{e^x}{e^x+1}\times\underbrace{\frac{e^x+1}{e^x+1}}_{=1}-\frac{e^x}{(e^x+1)^2}$

$=\frac{(e^x+1)^2-e^x(e^x+1)-e^x}{(e^x+1)^2}=\frac{e^{2x}+2e^x+1-e^{2x}-e^x-e^x}{(e^x+1)^2}=\frac{1}{(e^x+1)^2}$

Here we used the fact that $(e^{x})^2=e^{2x}$ . This is a general rule for real numbers: $(a^n)^m=a^{nm}$ .

Now, using this expansion and the fact that integration is linear (i.e. we can integrate term by term):

$I=\int \frac{1}{(e^x+1)^2}\,dx=\underbrace{\int 1\,dx}_{I_1}-\underbrace{\int\frac{e^x}{e^x+1}\,dx}_{I_2}-\underbrace{\int \frac{e^x}{(e^x+1)^2}\,dx}_{I_3}$

$I_1$

What function when differentiated gives $1$ ? The answer is $x+c$ : $I_1=x+c_1$ .

$I_2$

This is not in the tables. There is no obvious manipulation. Hence we may need a substitution. So looking for the function-derivative pattern, we see that $e^x+1$ has derivative $e^x$ so we let $u=$ “function”; $u=e^x+1$ (LIATE will also give you this – LIATE says takes the first thing on the list – the more complicated the better – and $e^x+1$ is more complicated that $e^x$ ):

$\frac{du}{dx}=e^x\Rightarrow dx= \frac{du}{e^x}$ .

Putting this back into $I_2$ :

$I_2=\int\frac{e^x}{u}\frac{du}{e^x}=\int\frac{du}{u}=\ln|u|+c$

$=\ln|e^x+1|+c_2$

$I_3$

Again there is no obvious rule or manipulation so we are looking for a substitution. Convince yourself that $u= e^x+1$ is the appropriate substitution…

$\frac{du}{dx}=e^x\Rightarrow dx=\frac{du}{e^x}$

Hence

$I_3=\int\frac{e^x}{u^2}\frac{du}{e^x}=\int u^{-2}\,du=\frac{u^{-1}}{-1}+c_3$

$=-\frac{1}{e^x+1}+c_3$

Now throwing this back into the original integral:

$I=I_1-I_2-I_3$

$=x+c_1-\ln|e^x+1|-c_2+\frac{1}{e^x+1}-c_3$

$=x+\frac{1}{e^x+1}-\ln|e^x+1|+c$

as $c_1-c_2-c_3$ is just some constant, which we can rename $c$ for short.

Question 8

Not in the tables and no obvious manipulation. Again, convince yourself (function-derivative pattern or LIATE) that $u=x^2+9$ is an appropriate substitution:

$\frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x}$ .

Putting back into the integral and suppressing the limits:

$I=\int \frac{x}{u}\frac{du}{2x}=\frac{1}{2}\int \frac{du}{u}$

$=\left[\frac{1}{2}\ln|u|\right]=\frac{1}{2}\left[\ln|x^2+9|\right]_0^3=\frac{1}{2}[\ln 18-\ln 9]$ .

Now one of the facts about logarithms is that:

$\log_a\left(\frac{x}{y}\right)=\log_ax-\log_ay$ ;

$I=\frac{1}{2}\ln\frac{18}{9}=\frac{\ln 2}{2}$ .

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MATH6037: Review of Integration Exercise Solutions

Question 1

(a)

(b)

(c)

Question 2

(a)

(b)

Question 3

Question 4

(a)

(b)

(c)

Question 5

Question 6

(a)

(b)

Question 7

$I_1$

$I_2$

$I_3$

Question 8

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MATH6037: Review of Integration Exercise Solutions

Question 1

(a)

(b)

(c)

Question 2

(a)

(b)

Question 3

Question 4

(a)

(b)

(c)

Question 5

Question 6

(a)

(b)

Question 7

Question 8

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