Evalute _ 1 3 ^1 (x-x^3 )^ 1 3 x^4 d x - Vidyadip

Question

Evalute $\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4} d x$

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Answer

$\therefore$ $I=\int_{1 / 3}^1 \frac{\left[x^3\left(\frac{1}{x^2}-1\right)\right]^{1 / 3}}{x^4} d x$
$\therefore$ $t=\int_{1 / 3}^1 \frac{x\left(\frac{1}{x^2}-1\right)^{1 / 3}}{x^4} d x$
$\therefore$ $=\int_{1 / 3}^1 \frac{\left(\frac{1}{x^2}-1\right)^{1 / 3}}{x^3} d x$
Let $\left(\frac{1}{x^2}-1\right)=t$
$\Longrightarrow-\frac{2}{x^3} d x=d t$
$\Longrightarrow \frac{d x}{x^3}=-\frac{d t}{2}$
Limits: $x=1 / 3 \rightarrow t=8, x=1 \rightarrow t=0$.
$I=\int_8^0 t^{1 / 3}\left(-\frac{d t}{2}\right)=\frac{1}{2} \int_0^8 t^{1 / 3} d t=\frac{1}{2}\left[\frac{3}{4} t^{4 / 3}\right]_0^8$
$I=\frac{3}{8}\left[8^{4 / 3}\right]=\frac{3}{8}[16]=6$

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