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Integration (1 Viewer)

Thread starter hantu_ah
Start date Feb 19, 2010

H

hantu_ah

New Member

#1

gurmies

Drover

#2

$\int_{0}^{\sin x}ydy=\frac{1}{2}\left [y^{2} \right ]^{\sin x}_{0}=\frac{1}{2}.\sin^{2}x \\\\ \therefore \int_{0}^{\pi}\int_{0}^{\sin x}ydydx=\frac{1}{2}\int_{0}^{\pi}\sin^{2}xdx \\\\ = \frac{1}{4}\int_{0}^{\pi}\left ( 1-\cos 2x \right )dx \\\\=\frac{1}{4}\left [x-\frac{1}{2}.\sin 2x \right ]^{\pi}_{0} \\\\ = \frac{\pi}{4}$

H

hantu_ah

New Member

#3

thanks

D

Drongoski

Well-Known Member

#4

Good work, Gurmies.

Kittikhun

Member

#5

At first look I would have given up right away but looking at the working out done by Gurmies it is pretty simple.

Last edited: Feb 23, 2010

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