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  • Choose the correct answer in Exercises 9 and 10. The value of the integral is

Choose the correct answer in Exercises 9 and 10. 

    Q9.    The value of the integral \int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx is

                (A)    6

                (B)    0

                (C)    3

                (D)    4

Answers (1)

best_answer

The value of integral is  (A) = 6

\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx
\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^\frac{1}{3}}{x^3}dx\\
let 
\frac{1}{x^2}-1 = t\Rightarrow \frac{dx}{x^3}=-dt/2
now, when x = 1/3, t = 8 and when x = 1 , t = 0

therefore

\\=-\frac{1}{2}\int_{8}^{0}t^{1/3}dt\\ =-\frac{1}{2}.\frac{3}{4}[t^4/3]^0_8\\ =-\frac{3}{8}[-2^4]\\ =6

Posted by

manish

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