Double Integrals!?

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

Double Integrals!?
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
Reply:I think that this question is not being stated properly ---please check and I will try again.
Reply:I am sorry but it makes non sense to have x vary between bounds which depend on x.
Reply:hahaha, I don't think I've done enough Calculus for this one...