Question

Question deleted.

Answer 1

let u = [64-(x-20)^2]
first integrate u^1/2 then multiply that by the integral of [64-(x-20)^2]

Answer 2

There are probably a few ways to approach this.
Personally, I would do a Trig sub.
\[\Large\rm \int\limits \sqrt{64-(x-20)^2}~dx\]Factoring a 64 out of each term gives us:\[\Large\rm 8\int\limits\sqrt{1-\left(\frac{x-20}{8}\right)^2}~dx\]Making the substitution:\[\Large\rm \frac{x-20}{8}=\sin \theta\]

Answer 3

\[\Large\rm dx=8\cos \theta~d \theta\]

Giving us an integral of something like...\[\Large\rm 8\int\limits \sqrt{1-\sin^2\theta}(8\cos \theta~d \theta)\]

Answer 4

How do I find theta?

Answer 5

Whut? +_+
Have you learned about Trigonometric Substitutions?

Answer 6

or have you solved ANY integral before ?

Answer 7

No i haven't learnt about trig substitutions.

I have solved integral before but without square roots...