int_0^2 frac{x}{(2-x)^{frac{3}{4}}} dx = | vedclass

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(C) Let $I = \int_0^2 \frac{x}{(2-x)^{3/4}} dx$. Substitute $t = 2-x$, then $dx = -dt$. When $x=0, t=2$ and when $x=2, t=0$. $I = \int_2^0 \frac{2-t}{

Vedclass · Accepted Answer

$\frac{32}{5} 2^{\frac{1}{4}}$

Vedclass · Answer

(C) Let $I = \int_0^2 \frac{x}{(2-x)^{3/4}} dx$. Substitute $t = 2-x$, then $dx = -dt$. When $x=0, t=2$ and when $x=2, t=0$. $I = \int_2^0 \frac{2-t}{t^{3/4}} (-dt) = \int_0^2 \frac{2-t}{t^{3/4}} dt$. $I = \int_0^2 (2t^{-3/4} - t^{1/4}) dt$. $I = [2 \cdot \frac{t^{1/4}}{1/4} - \frac{t^{5/4}}{5/4}]_0^2$. $I = [8t^{1/4} - \frac{4}{5}t^{5/4}]_0^2$. $I = 8(2^{1/4}) - \frac{4}{5}(2^{5/4}) = 8(2^{1/4}) - \frac{4}{5}(2 \cdot 2^{1/4})$. $I = 2^{1/4} (8 - \frac{8}{5}) = 2^{1/4} (\frac{40-8}{5}) = \frac{32}{5} 2^{1/4}$.

$\int_0^2 \frac{x}{(2-x)^{\frac{3}{4}}} dx = $

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