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).  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. 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:

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

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