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shayanreloaded

  • 2 years ago

Find the value of the integral

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  1. OpenSessame
    • 2 years ago
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    http://www.webmath.com/nintegrate.html Best website that gives you step by step instructions!

  2. zzr0ck3r
    • 2 years ago
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    \[\int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x-x^3}}{x^4}dx\]?

  3. shayanreloaded
    • 2 years ago
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    yes

  4. cinar
    • 2 years ago
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    \[ \Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x-x^3}}{x^4}dx\\ \Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{x^3(\frac{1}{x^2}-1)}}{x^4}dx\\ \Large \int_{\frac{1}{3}}^{1}\frac{x\sqrt[3]{\frac{1}{x^2}-1}}{x^4}dx\\ \Large \int_{\frac{1}{3}}^{1}\frac{\sqrt[3]{\frac{1}{x^2}-1}}{x^3}dx\\ \Large u=\frac{1}{x^2}=>du=\frac{-2}{x^3}dx\\ \Large \frac{-1}{2}\int\sqrt[3]{u-1}du\\ \]the reset is yours (: you can do one more subt.

  5. cinar
    • 2 years ago
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    I guess this subt. is better\[ \Large u=\frac{1}{x^2}-1=>du=\frac{-2}{x^3}dx\\ \Large \frac{-1}{2}\int\sqrt[3]{u}du\\ \Large \frac{-1}{2}\frac{3}{4}u^{\frac{4}{3}}\\ \Large \frac{-3}{8}(\frac{1}{x^2}-1)^{\frac{4}{3}}|_{\frac{1}3}^{1}=6\\ \]

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