Properties of definite integrals (1 Viewer)

[highkeydropout]

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

 

[highkeydropout]

The middle is another semicircle. Try subbing in u=x-3 for the middle

oh, do you mean after factorising it?

 

[highkeydropout]

yes

Okeyy ty

 

[highkeydropout]

so the final answer should be ​

I don’t get how you integrated the middle yet?

 

[tito981]

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]

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]

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!!

 

[CM_Tutor]

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: