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this question is driving me nuts, I can't think about an easy solution.

Let $F(x)=\int_{0}^{x} \sqrt{1+t^3}\,dt. $

Evaluate $\int_{0}^{2} x\,F(x)\,dx$ in terms of $F(2)$.

I know that the derivative of $F(x)$ is $\sqrt{1+x^3}$, but how is that supposed to help me?

Thanks!!

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    $\begingroup$ Integration by parts... $\endgroup$
    – copper.hat
    May 30, 2015 at 17:02
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    $\begingroup$ $xF(x)$ can be seen as $((d/dx)(0.5 x^2))F(x)$ and if you use integration by parts, you can get $F'(x)$ back $\endgroup$
    – Guest 86
    May 30, 2015 at 17:03
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    $\begingroup$ You should end up with something like $2F(2)-{26 \over 9}$. $\endgroup$
    – copper.hat
    May 30, 2015 at 17:06

1 Answer 1

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$$\int_{0}^{2}xF(x)dx=(IPP) \: 2F(2)-\frac{1}{2}\int_{0}^{2}x^{2}F'(x)dx\\ let \: \: I=\int_{0}^{2}x^{2}F'(x)dx=\int_{0}^{2}x^{2}\sqrt{1+x^{3}}dx\; \; : let \; \;u^{2}=x^{3} \rightarrow dx=\frac{2}{3}u^{-\frac{1}{3}}du\\ =\frac{2}{3}\int_{0}^{2^{\frac{3}{2}}}u\sqrt{1+u^{2}}du\; \; \; \; now\; \; let :u=sh(t)\rightarrow du=ch(t)dt\; \; , and: \sqrt{1+u^{2}}=ch(t)\\ finally :I=2\int_{0}^{sh(2^{\frac{3}{2}})}\frac{1}{3}sh(t)(ch(t))^{2}dt=2[(ch(t))^{3}]_{0}^{sh(2^{\frac{3}{2}})}$$

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