MATH6037: Approximate Integration Selected 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 p.87 of these notes.

Question 1

Estimate $\int_0^1\cos(x^2)\,dx$ using (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with $n=4$ .

Solution

Part (a)

For the Trapezoidal Rule, we have $\Delta x=0.25$ and we have the points $x_0=0$ , $x_1=0.25$ , $x_2=0.5$ , $x_3=0.75$ , $x_4=1$ . Hence using the formula:

$\int_0^1\cos(x^2)\,dx\approx\frac{0.25}{2}[\cos(0)+2[\cos(0.25^2)+\cos(0.5^2)+\cos(0.75^2)]$

$+\cos(1)]\approx 0.895795$ .

Part (b)

In this case we have $\Delta x=0.25$ and have midpoints $x_1=0.125$ , $x_2=0.375$ , $x_3=0.625$ and $x_4=0.875$ . Hence using the formula:

$\int_0^1\cos(x^2)\,dx\approx (0.25)[\cos(0.125^2)+\cos(0.375^2)+\cos(0.625^2)$

$+\cos(0.875^2)]\approx 0.908907$ .

Question 2 (b) (first integral only)

Use Simpson’s Rule to approximate to six decimal places.

$\int_0^\pi x^2\sin x\,dx$ , with $n=8$ .

Integrate by parts and compare these approximate values with the real value.

Not completed.

Solution

Now in this case $\Delta x=\pi/8$ and $x_i=i\pi/8$ . Hence using the formula:

$\int_0^\pi x^2\sin x\,dx\approx \frac{\pi/8}{3}[(0)^2\sin 0+4(\pi/8)^2\sin(\pi/8)+2(\pi/4)^2\sin(\pi/4)+$

$4(3\pi/8)^2\sin(3\pi/8)+2(\pi/2)^2\sin(\pi/2)+4(5\pi/8)^2\sin(5\pi/8)+$

$2(3\pi/4)^2\sin(3\pi/4)+4(7\pi/8)^2\sin(7\pi/8)+\pi^2\sin \pi]\approx 5.869247$ .

Now to integrate $I$ by parts choose $u=x^2$ by the LIATE rule; and thus $dv=\sin x\,dx$ . We have

$\frac{du}{dx}=2x\Rightarrow du=2x\,dx$ and $v=-\cos x$ .

Hence by the formula (we will put in limits later):

$I=x^2(\cos x)-\int \cos x(-2x\,dx)=-x^2\cos x+2\underbrace{\int x\cos x\,dx}_{:=J}$ .

Now to integrate $J$ we must integrate by parts a second time. Again by LIATE choose $u=x$ , thus $dv=\cos x\,dx$ . Hence

$\frac{du}{dx}=1\Rightarrow du=dx$ and $v=\sin x$ ;

$\Rightarrow J=x\sin x-\int\sin x\,dx=x\sin x+\cos x$ ,

$\Rightarrow I=-x^2\cos x+2x\sin x+2\cos x$ .

We must evaluate between the limits:

$\left[-x^2\cos x+2x\sin x+2\cos x\right]_0^\pi=-\pi^2\cos \pi+2\pi\sin\pi+2\cos \pi$

$-\left(-(0)^2\cos(0)+2(0)\sin 0+2\cos 0\right)=\pi^2-4\approx 0.8696014$ .

Hence, in this case, $E_S=-0.000357$ .

Question 4 (without implementation)

Estimate the errors involved in the approximations $T_8$ and $M_8$ for $\int_0^1 \cos(x^2)\,dx$ . How large do we have to choose $n$ so that the approximations $T_n$ and $M_n$ are accurate to within $0.00001$ ?

Solution

We have the upper bounds:

$|E_T|\leq\frac{K(b-a)^2}{12n^2}$ , and

$|E_M|\leq\frac{K(b-a)^3}{24n^2}$ .

where $K=\max_{x\in[a,b]}|f''(x)|$ . Now

$f'(x)=-\sin (x^2)(2x)$

$f''(x)=-\sin(x^2)(2)-2x\cos(x^2)(2x)=-2(\sin(x^2)+2x^2\cos(x^2))$ .

In terms of MATH6037 we haven’t gone into how to find the maximum of such a function. This question should not have been put in your exercises. In fact even the method I would use for solving this — namely the Closed Method http://129.81.170.14/~solson2/HO6.pdf (bottom of page 1) — has a great difficulty solving it. In this case I would leave $K$ as it is — as $K$ .

We know that $b-a=1$ and $n=8$ . Hence

$|E_T|\leq\frac{K}{768}$ , and

$|E_M|\leq \frac{K}{1536}.$

Again we haven’t got the tools to find $K$ but we can still find answers in terms of $K$ .

Now for the second part, looking at the Trapezoidal Rule first, we want to find an $n$ such that:

$|E_T|\leq 0.00001$ .

Now the maximum that $|E_T|$ can be is given by the formula so we can look at the inequality ( $b-a=1$ ):

$\frac{K(b-a)^3}{12n^2}<0.00001$ .

$\Rightarrow n^2>\frac{K}{12(0.00001)}=\frac{25000K}{3}$

$n>\sqrt{\frac{25000K}{3}}=50\sqrt{\frac{10K}{3}}$ .

Now looking at the Midpoint Rule, we want to find an $n$ such that:

$|E_M|<0.00001$ ,

so again we solve the inequality

$\frac{K(b-a^2)}{24n^2}<0.00001$

$\Rightarrow n^2>\frac{K}{24(0.00001)}=\frac{12500K}{3}$

$\Rightarrow n>\sqrt{\frac{12500K}{3}}=50\sqrt{\frac{5K}{3}}$ .

There is another, more elementary (than the closed interval method), option for estimating $K$ — although in our examples I’d like it to be obvious so this is just information really. The triangle inequality holds for all real numbers $x,y$ :

$|x+y|\leq |x|+|y|$ .

Hence, along with the fact that $|xy|=|x||y|$ and that $|x|=x$ for $x>0$ :

$|f''(x)|=|-2\sin (x^2)-4x^2\cos(x^2)|\leq 2|\sin(x^2)|+4x^2|\cos(x^2)|.$

Now $|\sin x|\leq 1$ and $|\cos x|\leq 1$ for all $x\in\mathbb{R}$ . Also $x^2$ has it’s maximum on $[0,1]$ at $x=1$ :

$|f''(x)|\leq 2+4x^2\leq 6$ .

Now $|f''(x)|\leq 6$ for all $x\in[0,1]$ so in particular the maximum of $|f''(x)|$ , namely $K$ will also be less than 6. So although it’s not as ‘sharp’ as possible we can still take $K=6$ and guarantee that our inequality still holds. Above I do it with a general $K$ .

This approximation of $K$ to 6 compares rather with the actual value of $K$ (found using Maple) of about $3.845$ .

Question 5 (a shortened version of)

Find the maximum error $|E_S|$ when we use the Simpson’s Rule with $n=10$ to approximate $\int_0^1 e^x\,dx$ . How large should $n$ be to guarantee that the approximations $S_n$ is accurate to $0.00001$ ?

Solution

Now we have the formula:

$|E_S|\leq \frac{K(b-a)^5}{180n^4}$ ,

where here $K=\max_{x\in[a,b]}|f^{(iv)}(x)|$ . We know that $b-a=1$ and that $n=10$ . All that remains to calculate is $K$ . Now as $f(x)=e^x$ , $f^{(iv)}(x)=e^x$ . Now $e^x$ is an increasing function of $x$ , hence it’s maximum is found when $x$ is large as possible — in this case at $1$ . Hence $K=e^1=e\approx 2.71828$ . Hence

$|E_S|\leq \frac{e}{180(8)^4}\approx 0.000003$ .

How large to get closer that $0.00001$ — well clearly less than (or equal ;-)) $8$ in the case of Simpson’s Rule. We need to solve the inequality for $n$ :

$\frac{K(b-a)^5}{180n^4}<0.00001$

$\Rightarrow n^4>\frac{e}{180(0.00001)}=\frac{5000e}{9}$

$n>\sqrt[4]{\frac{5000e}{9}}\approx 6.2$ .

Now because the Simpson’s Rule is for even $n$ , that $n\geq 8$ is both necessary and sufficient for the error to be less than $0.00001$ ; i.e. the answer is $8$ .

Question 7*

Show that if $p$ is a polynomial of degree 3 or lower, then Simpson’s Rule gives the exact value of $\int_a^b p(x)\,dx$ .

Solution

Let $p(x)=ax^3+bx^2+cx+d$ with $a,b,c,d\in\mathbb{R}$ . Now

$p'(x)=3ax^2+2bx+c$

$p''(x)=6ax+2b$

$p'''(x)=6a$

$p''''(x)=0$ .

Now if the fourth derivative of $p$ is 0 for all $x\in\mathbb{R}$ , then for any interval $[a,b]$ , $\max_{x\in[a,b]}|f^{(iv)}(x)|=0$ ; hence $K=0$ and so for any $n\in\mathbb{N}$ :

$|E_S|=\frac{K(b-a)}{180n^4}=0$ ,

that is the error in Simpson’s Rule is zero — that is Simpson’s Rule gives the exact value of the integral $\bullet$

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MATH6037: Approximate Integration Selected Exercise Solutions

Question 1

Solution

Part (a)

Part (b)

Question 2 (b) (first integral only)

Solution

Question 4 (without implementation)

Solution

Question 5 (a shortened version of)

Solution

Question 7*

Solution

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MATH6037: Approximate Integration Selected Exercise Solutions

Question 1

Solution

Part (a)

Part (b)

Question 2 (b) (first integral only)

Solution

Question 4 (without implementation)

Solution

Question 5 (a shortened version of)

Solution

Question 7*

Solution

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