Question

How do I find the integral from -4 to 4 of (3*sqrt(16-x^2)^2 dx ?

I think you have to use u-substitution but I don't entirely understand how that works in this case.

Answer 1

use nothing

Answer 2

well the sqrt gets eliminated with the power 2, so...Its a simple integral

Answer 3

\[y=\sqrt{16-x^2}\] is the upper half of a circle centered at the origin with radius 4
so
\[6\int_{-4}^4\sqrt{16-x^2}dx\] is 6 times the area of the upper half of a circle with radius 4

Answer 4

oh sorry i didn't see the square at the end
if it is really the square root squared, then it is easy enough, although a rather strange way to write it

Answer 5

it is really
\[3\int_{-4}^4\sqrt{16-x^2}^2dx\]?

Answer 6

Assuming it's what sattelite said then this becomes very simple.

Answer 7

Then this become easy to integrate.