I'm having trouble calculating the integral of $\sqrt{X}(1-x)$ with substitution.

The steps which I have so far:

$$\sqrt{x} (1-x)dx \quad \{u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}\}$$

$$2 \int -u^2(u^{2}-1)du$$

$$-2 \int u^2(u^{2}-1)du$$

Could someone please help me calculate this integral using substitution?

  • $\begingroup$ Maybe you get more answers if you have a look at the MathJax tutorial $\endgroup$ – Miguel Oct 1 '17 at 19:44
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    $\begingroup$ After substitution, you can expand the polynomial and integration is immediate. $\endgroup$ – Miguel Oct 1 '17 at 19:47
  • $\begingroup$ I'm not entirely sure what the $f$ is supposed to be, but if it's supposed to be an integral, please use \int. $\endgroup$ – Simply Beautiful Art Oct 1 '17 at 19:49
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    $\begingroup$ As Miguel says, just distribute to get$$u^2(u^2-1)=u^4-u^2$$and apply the power rule. $\endgroup$ – Simply Beautiful Art Oct 1 '17 at 19:52
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    $\begingroup$ The integrand is $x^{1/2}-x^{3/2}$ and can be integrated much more quickly without substitution. $\endgroup$ – Aretino Oct 1 '17 at 19:52

Well, substitution isn't much helpful here if you know the anti derivative of ${x}^n $. But still:

Let, $\sqrt{x}=p\implies dx=2p\cdot dp$

$$\int{\sqrt{x}(1-x)}\cdot dx=\int{p(1-p^2)2p\cdot dp}=-2\int{p^4-p^2}=\frac{2p^3}{3}-\frac{2p^5}{5}=\frac{2x\sqrt{x}}{3}-\frac{2x^2\sqrt{x}}{3}$$


You don't need any substitution!

$$\int (1-x)\sqrt{x}\,dx=\int (x^{1/2}-x^{3/2})\,dx=(*)$$

Now remember that, for every $\alpha\ne-1$; $$\int x^\alpha\,dx=\frac{1}{1+\alpha}\,x^{1+\alpha}+C$$

So we have

$$(*)=\dfrac{1}{1+\frac12}\,x^{1+\frac12}-\dfrac{1}{1+\frac32}\,x^{1+\frac32}+C=\dfrac{2}{3}\,x^{\frac32}-\dfrac{2}{5}\,x^{\frac52}+C$$ hence

$$(*)=\dfrac{2}{3}x\sqrt{x}-\dfrac{2}{5}x^2\sqrt{x}+C=\dfrac{2}{15}x\sqrt{x}\left(5-3x \right)+C$$

  • $\begingroup$ I agree that you don't need substitution but I'm required to $\endgroup$ – Anna Jeanine Oct 1 '17 at 20:03
  • $\begingroup$ Hope it can be useful anyway. I do math not homeworks :) $\endgroup$ – Raffaele Oct 1 '17 at 20:05
  • $\begingroup$ It's not homework, it's trying to get better at math ;-) $\endgroup$ – Anna Jeanine Oct 1 '17 at 20:07
  • $\begingroup$ That's wonderful! You'll appreciate my simple and elegant and modest solution ... :)) $\endgroup$ – Raffaele Oct 1 '17 at 20:21
  • $\begingroup$ Please learn to use $$...$$ instead of $...$ and \frac instead of \dfrac (which is very rarely needed). $\endgroup$ – Did Oct 2 '17 at 6:15

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