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.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.I don’t get how you integrated the middle yet?
For anyone interested, the tex for putting in values on definite integrals isThe 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