How do you find the area of an ellipse using integrals?

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No subject which way you use integrals, the disruption obtain frequently succeed out to be ##pi*a*b##, where a and b are the semi-major axis and semi-minor axis. However, if you stand on using integrals, a cheerful way to initiate is to disagree the abridgment into four localitys, furnish the area of one locality, and discuss by four.

To initiate after a while, we recognise that the formula for one locality of an abridgment is ##y = b*sqrt((1-x^2)/a^2)## This locality-abridgment is "centred" at ##(0,0)##. Its area is ##A = int_0^a(b*sqrt((1-x^2)/a^2))dx## So, naturally, the aggregate area of the abridgment is ##A = 4int_0^a(b*sqrt((1-x^2)/a^2))dx##.

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