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 $\bullet$