Integration (1 Viewer)

clry · Aug 6, 2011

Integration mind blank - HELP!

Find area enclosed between the curve y = 2 / (x-3)^2 , the x-axis, and the lines x = 0 and x = 1

I just get confused when I make it 2(x-3)^-2 and I try to differentiate that, what happens to the 2?

SpiralFlex · Aug 6, 2011

$\int_{0}^{1}2(x-3)^-^2 \: dx$

$=-2(x-3)^-^1$ Limits: 0 to 1.

$=-\frac{2}{(x-3)}$ Limits: 0 to 1.

$=\frac{1}{3}$ units squared.

clry · Aug 6, 2011

Cheers

didn't realise we just left the 2 as is out front!

Integration (1 Viewer)

clry

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SpiralFlex

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clry

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