int_0^1 frac{x}{(1-x)^{3/4}} dx = | vedclass

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(D) Let $I = \int_0^1 \frac{x}{(1-x)^{3/4}} dx$. Substitute $t = 1 - x$,then $dt = -dx$. When $x = 0, t = 1$ and when $x = 1, t = 0$. $I = \int_1^0 \f

Vedclass · Accepted Answer

$\frac{16}{5}$

Vedclass · Answer

(D) Let $I = \int_0^1 \frac{x}{(1-x)^{3/4}} dx$. Substitute $t = 1 - x$,then $dt = -dx$. When $x = 0, t = 1$ and when $x = 1, t = 0$. $I = \int_1^0 \frac{1-t}{t^{3/4}} (-dt) = \int_0^1 \frac{1-t}{t^{3/4}} dt$. $I = \int_0^1 (t^{-3/4} - t^{1/4}) dt$. $I = \left[ \frac{t^{1/4}}{1/4} - \frac{t^{5/4}}{5/4} \right]_0^1$. $I = \left[ 4t^{1/4} - \frac{4}{5}t^{5/4} \right]_0^1 = 4 - \frac{4}{5} = \frac{20-4}{5} = \frac{16}{5}$.

$\int_0^1 \frac{x}{(1-x)^{3/4}} dx = $

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