Double Integrals!?

Question

The question in an exam paper asks:
The value of the double integral:
Integral, Upper limit = x, Lower limit = x/3, 2nd integral, Upper limit = 4sinx/2, lower limit = 0, (dydx).

Thats it there is no Function to integrate. Just the dydx after the two integrals.
It gives multiple choice answers of:
A = -4sqrt3
B= 5sqrt2
C = 2sqrt3
D=4sqrt3
E = None of these

Answer 1

Charlie,

I am afraid that you HAVE copied the question down wrongly...
...but I think I've sussed it anyway.

Because there is no explicit integrand that means that the integrand is actually 1.

You have given the outer limits as x and x/3. In fact, it must be pi and pi/3.

Let S stand for the integral sign and pretent that the limits are in place:

SS 1 dydx = S [y] dx = S (4sin(x/2) - 0) dx = S 4sin(x/2) dx

This performs the 'inner' integral (the y-integral). Remember, I have not written the limits in until the calculation - you have to pretend they are there on the integral sign and the square brackets.

Then,

S 4sin(x/2) dx = [-8cos(x/2)] = -8cos(pi/2) - -8cos(pi/6) = 8 sqrt(3)/2 = 4sqrt(3) because cos(pi/2) = 0 and cos(pi/6) = sqrt(3)/2

So the correct answer is D.

Hope this makes sense!

Perspykashus

Answer 2

I am sorry but it makes non sense to have x vary between bounds which depend on x.

Answer 3

I think that this question is not being stated properly ---please check and I will try again.

Answer 4

hahaha, I don't think I've done enough Calculus for this one...