how to do this integration question please. urgent (1 Viewer)

girlworld_club · May 5, 2013

http://tinypic.com/view.php?pic=30dcj0y&s=5

the question asks to evaluate.

girlworld_club · May 5, 2013

Please show full working.

Registered User · May 5, 2013

If you can't do it straight away just use a substitution. Let u=3-2x and let u=x-2.

pokka · May 5, 2013

$\int_0^2(3-2x)^5 dx\\\\=\left[\dfrac{(3-2x)^6}{6\times -2}\right]^2_0\\\\=\left[\dfrac{(3-2x)^6}{6\times -2}\right]^2_0$
Then substitute from there. Similar method for the second question except the power is 1/2.

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how to do this integration question please. urgent (1 Viewer)

girlworld_club

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