## Test 2 Solutions

Please find the test 2 solutions and marking scheme here.

## Exam Layout

Answer Q.1 [40 Marks] and two out of Q. 2, 3, 4 [30 marks each].

Q. 1 — 8 short questions each worth 5 marks each

Q. 2 — Differentiation with applications

Q. 3 — Integration with applications

Q.4 — Applications of differentiation and integration

## 2 comments

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December 14, 2012 at 3:15 pm

James GogginI was wondering if you could give me a hand with a couple of questions from the 2009 exam paper Q 1 d: given that , evaluate and

and Q 1 e:

.

Cheers

December 14, 2012 at 4:38 pm

J.P. McCarthyJames,

means the derivative of . Note also that

‘the derivative of

Here means find the value of the derivative when .

is the product of the functions and . The Product Rule says the derivative of a product , rather than is

.

In words, the first function by the derivative of the second, plus the second function times the derivative of the first. We can get the derivative very easily: . The derivative of is actually presented in the tables as

for a constant .

This comes from a Chain Rule:

.

All in all we have

.

Now evaluate at :

.

Similarly

.

Now for your second question. The first thing is to say is that integration is linear. These means that you can pull out constants and differentiate a sum of terms term-by-term. This means that what we are looking at is

.

Can this be done directly? I would argue no because

and

.

are not in the tables where is a constant. What about a manipulation? Well there are many but none of them actually simplify the integrand. This means that we need a substitution. Say for the first one we could pick . We could do this because

is a function which has derivative which is just a ‘multiple’…

LIATE — no logs, no inverse trig, algebraic; yes .

is ‘inside’ another function.

So we implement the substitution

.

Now put the integral back together

.

The integral of with respect to is — as the derivative of with respect to is . Hence we write

.

Similarly,

.

Some people learn these formula by comparing with

.

We will do that hear and say that these integrals can be integrated directly:

There are ways to evaluate the sines and cosines without a calculator but at this late date I either assume that you know how to do this or will use the calculator. I warn you that if you are using the calculator make sure that you are in radian mode:

Regards and Happy Christmas,

J.P.