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**1. find the antiderivative of sqrt(81-x^2) in the bounds 0 to 9/2**

okay so the first thing i did was sin substitution.

where I made x = 9sin(t) dx = 9cos(t) where t is theta.

then after some manipulating i got that that new integral was 9cos^2(t)

i then used the double angle formula and made it equal to 9/2 * (1 + cos(2t))

I then found the antiderivative of this integral to be 9/2 * (t + sin(2t)/2)

Using the identity that sin(2t) = 2sin(t)cos(t) the antiderivative became

(9/2) (t + cos(t)sin(t))

I then changed the bounds of the integral.

When x goes from 0 to 9/2, theta goes from 0 to pi/6

i then solved the integral to get 4.304

The actual answer is 38.74 so i think im completely missing the boat here.