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The value of the integral $\int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x$ is
(a) 6
(b) 0
(c) 3
(d) 4
The value of the integral $\int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x$ is
(a) 6
(b) 0
(c) 3
(d) 4
Solution:
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Verified Answer
(a) Let $\mathrm{I}=\int_{1 / 3}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x=\int_{1 / 3}^1 \frac{x^{1 / 3}\left(1-x^2\right)^{1 / 3}}{x^4} d x$
Let $1-x^2=t^3 \Rightarrow x^2=1-t^3 \Rightarrow-2 x d x=3 t^2 d t$
Consider $\mathrm{I}=-\frac{1}{2} \int \frac{x^{1 / 3}\left(1-x^2\right)^{1 / 3}(-2 x)}{x^4 \cdot x} d x$
$\left.\mathrm{I}=-\frac{1}{2} \int \frac{\left(1-t^3\right)^{1 / 3} t}{\left(1-t^3\right)^2} 3 t^2 d t=\frac{-3}{4(x)^2} \therefore \mathrm{I}=\frac{-3}{4} x^2\right]_{\frac{1}{3}}^1=6$
Let $1-x^2=t^3 \Rightarrow x^2=1-t^3 \Rightarrow-2 x d x=3 t^2 d t$
Consider $\mathrm{I}=-\frac{1}{2} \int \frac{x^{1 / 3}\left(1-x^2\right)^{1 / 3}(-2 x)}{x^4 \cdot x} d x$
$\left.\mathrm{I}=-\frac{1}{2} \int \frac{\left(1-t^3\right)^{1 / 3} t}{\left(1-t^3\right)^2} 3 t^2 d t=\frac{-3}{4(x)^2} \therefore \mathrm{I}=\frac{-3}{4} x^2\right]_{\frac{1}{3}}^1=6$
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