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from 0->1:Int sqrt(1-x^2) Solution)
I=∫sqrt(1-x2)dx = sqrt(1-x2)∫dx - ∫{(-2x)/2sqrt(1-x2)}∫dx ---->(INTEGRATION BY PARTS) = x√(1-x2) - ∫-x2/√(1-x2)
Let I1= ∫x2/√(1-x2) = ∫1-x2-1/√(1-x2) = ∫√(1-x2) - ∫1/√(1-x2) = ∫√(1-x2) - sin-1x = I - sin-1x
I = x√(1-x2) - I + sin-1x
2I = x√(1-x2) + sin-1x
I = x/2*√(1-x2) + 1/2*sin-1x
Within limits 0 -> 1
I = [1/2*√(1-1) + 1/2sin-11 - 1/2*√(1-0) - 1/2sin^-1(0) ] = 0 + pi/2 - 1/2 - 0 = pi/2 - 1/2
if P is a point in the interior of a triangles ABC,prove that AB>BC+CA
All about fractions
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