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 1x^{-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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