# Thread: How do l split this integral ?

1. ## How do l split this integral ?

• $\displaystyle pg\int_0^2 2\sqrt{4- x^2}(5+x)dx$

Is there a way to split this integral ?

2. Hello,
Originally Posted by nyasha

• $\displaystyle pg\int_0^2 2\sqrt{4- x^2}(5+x)dx$

Is there a way to split this integral ?

$\displaystyle \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx$
For the first one, substitute x=2sin(u)
For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)

3. Originally Posted by Moo
Hello,

$\displaystyle \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx$
For the first one, substitute x=2sin(u)
For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)
For the first integral can l substitute x=Rsinθ ?

4. Originally Posted by nyasha
For the first integral can l substitute x=Rsinθ ?
that's what I meant... but R is such that Rē=4, so R=2

5. Originally Posted by Moo
that's what I meant... but R is such that Rē=4, so R=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 ?

6. That thing does not go to zero.

7. Originally Posted by nyasha
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 $\displaystyle \int_0^1 x^2 - x \, dx \neq 0$.

8. Originally Posted by Moo
Hello,

$\displaystyle \int_0^2 \sqrt{4-x^2}(5+x) ~dx=\int_0^2 5 \sqrt{4-x^2} ~dx+\int_0^2 x \sqrt{4-x^2} ~dx$
For the first one, substitute x=2sin(u)
For the second one, substitute t=sqrt(4-x^2) (or just t=4-x^2)
Thanks