# 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

which is a semicircle. Specifically the area of this is I think
PS pretend the square brackets are the bounds

Last edited:

#### 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

#### highkeydropout

##### Member
so the final answer should be
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.

• Sunaina123 and highkeydropout

#### 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 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 units squared or a quarter of the circle's area.
yup tysm!!

• notme123

#### CM_Tutor

##### Moderator
Moderator
The middle is another semicircle. Try subbing in u=x-3 for the middle.
This way you'll get

which is a semicircle. Specifically the area of this is 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: