Question

(Definite Intergral) from 0 to 4 (rad: 16-x^2) dx by using trig functions

Answer 1

\[\int\limits_{0}^{4}(\sqrt{16-x^2}) dx\] yes?

Answer 2

\[\int\limits_{0}^{4} \sqrt{16-x^2} dx\] \[=8\sin^{-1} (x/4) + (x \sqrt{16-x^2})/2 \left| \right|_{0}^{4}\]\[=4\pi\]