int_{frac{1}{3}}^1 frac{(x-x^3)^{frac{1}{3}}} | vedclass

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(D) Let $I = \int_{\frac{1}{3}}^1 \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx$. Factor out $x^3$ from the term inside the cube root: $I = \int_{\frac{1}{3}}^

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Let $I = \int_{\frac{1}{3}}^1 \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx$. Factor out $x^3$ from the term inside the cube root: $I = \int_{\frac{1}{3}}^1 \frac{(x^3(\frac{1}{x^2}-1))^{\frac{1}{3}}}{x^4} dx = \int_{\frac{1}{3}}^1 \frac{x(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^4} dx = \int_{\frac{1}{3}}^1 \frac{(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^3} dx$. Let $u = \frac{1}{x^2} - 1$. Then $du = -\frac{2}{x^3} dx$,which implies $\frac{dx}{x^3} = -\frac{1}{2} du$. When $x = \frac{1}{3}$,$u = \frac{1}{(1/3)^2} - 1 = 9 - 1 = 8$. When $x = 1$,$u = \frac{1}{1^2} - 1 = 0$. Substituting these into the integral: $I = \int_8^0 u^{\frac{1}{3}} (-\frac{1}{2}) du = \frac{1}{2} \int_0^8 u^{\frac{1}{3}} du$. $I = \frac{1}{2} [\frac{u^{\frac{4}{3}}}{4/3}]_0^8 = \frac{1}{2} \cdot \frac{3}{4} [u^{\frac{4}{3}}]_0^8 = \frac{3}{8} (8^{\frac{4}{3}} - 0) = \frac{3}{8} (16) = 6$. Thus,the correct option is $D$.

$\int_{\frac{1}{3}}^1 \frac{(x-x^3)^{\frac{1}{3}}}{x^4} dx = $ . . . . . . .

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