\(\int_{\frac{1}{3}}^{1}\displaystyle\frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4}\,dx=\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6
Taking \(x^3\) out of the bracket, \(x - x^3 = x^3\left(\frac{1}{x^2} - 1\right)\), so
\( \displaystyle \left(x-x^3\right)^{\frac{1}{3}} = x\left(\frac{1}{x^2}-1\right)^{\frac{1}{3}} \)
\( \displaystyle I = \int_{\frac{1}{3}}^{1} \frac{1}{x^3}\left(\frac{1}{x^2}-1\right)^{\frac{1}{3}}dx \)
Let \(t = \frac{1}{x^2} - 1\). Then \(dt = -\frac{2}{x^3}\,dx\), so \(\frac{dx}{x^3} = -\frac{dt}{2}\). Changing the limits:
\( \displaystyle x = \frac{1}{3}:\ t = 9 - 1 = 8 \)
\( x = 1:\ t = 1 - 1 = 0 \)
\( \displaystyle I = -\frac{1}{2}\int_{8}^{0} t^{\frac{1}{3}}\,dt = \frac{1}{2}\int_{0}^{8} t^{\frac{1}{3}}\,dt \)
\( \displaystyle = \frac{1}{2}\left[\frac{3}{4}t^{\frac{4}{3}}\right]_{0}^{8} \)
\( \displaystyle = \frac{3}{8}\times 8^{\frac{4}{3}} = \frac{3}{8}\times 16 = 6 \)