# Properties of definite integrals (1 Viewer)

#### highkeydropout

##### Member
hey guys, not too sure how to solve the equation (specifically the middle). Thank you

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#### notme123

##### Member
The middle is another semicircle. Try subbing in u=x-3 for the middle.
This way you'll get $\bg_white \int[0][3]{{\sqrt{(3+u)(3-u)}du$
$\bg_white =\int[0][3]{{\sqrt{9-u^2}du$
which is a semicircle. Specifically the area of this is $\bg_white \frac{9}{4} pi$ I think
PS pretend the square brackets are the bounds

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#### highkeydropout

##### Member
The middle is another semicircle. Try subbing in u=x-3 for the middle
oh, do you mean after factorising it?

Okeyy ty

#### notme123

##### Member
so the final answer should be $\bg_white \frac{-5pi}{3} + \frac{\sqrt{3}}{2$

#### highkeydropout

##### Member
so the final answer should be $\bg_white \frac{-5pi}{3} + \frac{\sqrt{3}}{2$
I don’t get how you integrated the middle yet?

#### tito981

##### Active Member
I don’t get how you integrated the middle yet?
He didnt integrate, he realized it was a semi circle and after a sub and then he used the area of a circle formula to find the area beneath the curve. This is just an instance where it is easier to use properties of the curve rather than integrating it.

#### notme123

##### Member
I don’t get how you integrated the middle yet?
sorry for the late reply, like tito said I subbed in u=x-3 and the equation turned into a semicircle, the area from u=0 to the edge of the semicircle (u=3). so the area of the integral is $\bg_white \frac{\pi(3)^2}{4}$ units squared or a quarter of the circle's area.

#### highkeydropout

##### Member
sorry for the late reply, like tito said I subbed in u=x-3 and the equation turned into a semicircle, the area from u=0 to the edge of the semicircle (u=3). so the area of the integral is $\bg_white \frac{\pi(3)^2}{4}$ units squared or a quarter of the circle's area.
yup tysm!!

#### CM_Tutor

##### Moderator
Moderator
The middle is another semicircle. Try subbing in u=x-3 for the middle.
This way you'll get $\bg_white \int[0][3]{{\sqrt{(3+u)(3-u)}du$
$\bg_white =\int[0][3]{{\sqrt{9-u^2}du$
which is a semicircle. Specifically the area of this is $\bg_white \frac{9}{4} pi$ I think
PS pretend the square brackets are the bounds
For anyone interested, the tex for putting in values on definite integrals is

\int_{min value}^{max value}

and Greek letters are written with a slash, ie:

\pi

So, the above becomes:

$\bg_white \int_{0}^{3}{{\sqrt{(3+u)(3-u)}\ du=\int_{0}^{3}{{\sqrt{9-u^2}\ du=\frac{9\pi}{4}$