*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 using (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with .*

## Solution

### Part (a)

For the Trapezoidal Rule, we have and we have the points , , , , . Hence using the formula:

.

### Part (b)

In this case we have and have midpoints , , and . Hence using the formula:

.

## Question 2 (b) (first integral only)

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

*, with .*

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

*Not completed.*

### Solution

Now in this case and . Hence using the formula:

.

Now to integrate by parts choose by the LIATE rule; and thus . We have

and .

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

.

Now to integrate we must integrate by parts a second time. Again by LIATE choose , thus . Hence

and ;

,

.

We must evaluate between the limits:

.

Hence, in this case, .

## Question 4 (without implementation)

*Estimate the errors involved in the approximations ** and for . ** How large do we have to choose so that the approximations and are accurate to within ?*

### Solution

We have the upper bounds:

, and

.

where . Now

.

*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 as it is — as .*

We know that and . Hence

, and

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

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

.

Now the maximum that can be is given by the formula so we can look at the inequality ():

.

.

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

,

so again we solve the inequality

.

*There is another, more elementary (than the closed interval method), option for estimating — 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 :*

*.*

*Hence, along with the fact that and that for :*

*Now and for all . Also has it’s maximum on at :*

*.*

*Now for all so in particular the maximum of , namely will also be less than 6. So although it’s not as ‘sharp’ as possible we can still take ** and guarantee that our inequality still holds. Above I do it with a general .*

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

## Question 5 (a shortened version of)

*Find the maximum error when we use the Simpson’s Rule with to approximate . How large should be to guarantee that the approximations is accurate to ?*

### Solution

Now we have the formula:

,

where here . We know that and that . All that remains to calculate is . Now as , . Now is an increasing function of , hence it’s maximum is found when is large as possible — in this case at . Hence . Hence

.

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

.

Now because the Simpson’s Rule is for *even , *that is both necessary *and *sufficient for the error to be less than ; i.e. the answer is .

## Question 7*

*Show that if is a polynomial of degree 3 or lower, then Simpson’s Rule gives the exact value of .*

### Solution

Let with . Now

.

Now if the fourth derivative of is 0 for all , then for any interval , ; hence and so for any :

,

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

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