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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
x=ct,y=c/t d^2/dx^2
#question if two angles of a triangle are unequal in measure then the side opposite to greater angle is longer than the side opposite to the smaller angle
HOW MANY ZERO ARE THERE AT THE END OF 200
sin10+sin20+sin30+....+sin360=0 sin10+sin20+sin30+sin40+...sin180+sin(360-170)+......+sin(360-40)+sin(360-30)+sin(360-20)+sin360-10)+sin360 sin360-x=-sinx hence all terms cancel
Find out the volume of the solid obtained by rotating the region bounded by y = x 2 - 4x + 5 , x = 1 , x = 4 , and the x-axis about the x-axis. Solution : The firstly thing t
4n to the power 3/2 = 8 to the power minus 1/3. find the value of n.
usefullness of product life cycle
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I need help witth my homework can you help please
Continuity : In the last few sections we've been using the term "nice enough" to describe those functions which we could evaluate limits by just evaluating the function at the po
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