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Evaluate $\int\limits_0^2 x^4(8-x^3)^{\dfrac{-1}3}$

written 10.0 years ago by teamques10 ★ 70k

modified 4.4 years ago by krithikkm200 • 10

Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 3

Year : 2014

engineering mathematics

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written 10.0 years ago by teamques10 ★ 70k

•   modified 10.0 years ago

Put $x^3=8t\space \space \therefore x=2t^{\dfrac13} $

$dx=\dfrac 23t^{-\dfrac23}dt$

When $x=2\space \space t=1$

$X=0\space \space t=0$

$I=\dfrac{16}3\int\limits_0^1t^{-3}(1-t)^{\dfrac{-1}3}$dt

$I=\dfrac {16}3B\Bigg(\dfrac 53,\dfrac 23\Bigg)$

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