$\displaystyle \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx$
For the first one, substitute x=2sin(u)
For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)
$\displaystyle \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx$
For the first one, substitute x=2sin(u)
For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)
Will l be correct if l say this integral$\displaystyle \int_0^2 x \sqrt{4-x^2} ~dx$ is zero so l can try to ignore it ?
$\displaystyle f(a) = f(b) = 0$ does not imply that $\displaystyle \int_a^b f(x) \, dx = 0$ eg. It should be obvious that $\displaystyle \int_0^1 x^2 - x \, dx \neq 0$.
$\displaystyle \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx$
For the first one, substitute x=2sin(u)
For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)