problem with integration?

Question

I have been asked to find th value of:

int:(4-x^2)^(1/2)dx with limits from x=0 to x=1 and using x=2sin(theta) as a substitution.

I have come up with the answer of 0.181, is this correct. My integration gives me (-sin2(theta)+2(theta)) with limits of theta=0 and theta =pi/6. If this is wrong, why.

I've just realised the integration should become sin2(theta)+2(theta) not -sin2(theta)+2(theta)

My new answer is 1.913

Answer 1

x = 2sin(theta)
dx/d(theta) = 2cos(theta)
dx = 2cos(theta).d(theta)

Also, when x=0 and 1, theta=0 and 30 degrees

int:(4-x^2)^(1/2)dx
= int:(4-4sin^2(theta))^(1/2)[2cos(theta)]d(theta)
= int:[4(1-sin^2(theta))]^(1/2)[2cos(theta)]d(theta)
= int:[2cos(theta)][2cos(theta)]d(theta)
= int:2(2cos^2(theta))d(theta)

Since cos2(theta) = 2cos^2(theta) - 1, so
int:2(2cos^2(theta))d(theta)
= int: [2(cos2(theta) + 1]d(theta)
= int: [2cos2(theta) + 2] d(theta)
= [2sin2(theta)]/2 + 2(theta)] from 0 to 30 (or pi/6)
= (sin60 + pi/3) - (sin0 + 0)
= (SQRT3)/2 + pi/3
= 1.91 (to 3 sig fig)

Answer 2

Since there could be problems with range of the angle which is actually rare, but it is always advisable to change resubstitute into the original and then put limits to get the answer.

So, ur intergration looks fine.

Answer 3

Offhand, it looks good to me. I'll keep that substitution in mind. Just make sure you back-substitute correctly to x from theta.

Answer 4

I don't think the answer is right? f(0) = 2 and f(1) = 1.732 so this suggests an area between 2 and 1.732