Question

Find the value of the integral

Answer 1

http://www.webmath.com/nintegrate.html
Best website that gives you step by step instructions!

Answer 2

\[\int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x-x^3}}{x^4}dx\]?

Answer 3

yes

Answer 4

\[ \Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x-x^3}}{x^4}dx\\
\Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x^3(\frac{1}{x^2}-1)}}{x^4}dx\\

\Large \int_{\frac{1}{3}}^{1}\frac{x\sqrt[3]{\frac{1}{x^2}-1}}{x^4}dx\\
\Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{\frac{1}{x^2}-1}}{x^3}dx\\

\Large u=\frac{1}{x^2}=>du=\frac{-2}{x^3}dx\\

\Large \frac{-1}{2}\int\sqrt[3]{u-1}du\\



\]the reset is yours (: you can do one more subt.

Answer 5

I guess this subt. is better\[
\Large u=\frac{1}{x^2}-1=>du=\frac{-2}{x^3}dx\\

\Large \frac{-1}{2}\int\sqrt[3]{u}du\\
\Large \frac{-1}{2}\frac{3}{4}u^{\frac{4}{3}}\\

\Large \frac{-3}{8}(\frac{1}{x^2}-1)^{\frac{4}{3}}|_{\frac{1}3}^{1}=6\\
\]