Properties of definite integrals (1 Viewer)

notme123

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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
 
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tito981

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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

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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.
 
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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

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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:

 

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