1
$\begingroup$

I have to find $$\int\limits_{0}^{1}x(x+a)^{1/p}\text{ d}x$$ and I'm told to use two substitutions, $u=(x+a)^{1/p}$ and $u=x+a$.

However, I can't find the derivative of $u$, $\text{d}u$, to replace once I've done the substitution - could someone offer a bit of insight into what I'm missing here? Thank you!

$\endgroup$
  • 2
    $\begingroup$ Use the power rule and the chain rule. $\endgroup$ – TorsionSquid Sep 6 '15 at 19:22
  • 6
    $\begingroup$ Easier, I think, to write $x=(x+a)-a$ and split the integral into two. $\endgroup$ – lulu Sep 6 '15 at 19:23
  • $\begingroup$ For the substitution $u=x+a$, you can use $du=dx$ (since $\frac{du}{dx}=1$). For the substitution $u=(x+a)^{1/p}$, you have $x=u^p-a$ so $dx=pu^{p-1}du$. $\endgroup$ – user84413 Sep 6 '15 at 22:32
3
$\begingroup$

Hint: $$\int_{0}^{1}x(x+a)^{1/p}\,dx = \int_{0}^{1}(x+a)^{1+1/p}\,dx-a\int_{0}^{1}(x+a)^{1/p}\,dx$$ and: $$\int_{0}^{1}(x+a)^{\beta}\,dx = \left.\frac{(x+a)^{\beta+1}}{\beta+1}\right|_{0}^{1}=\frac{(1+a)^{\beta+1}-a^{\beta+1}}{1+\beta}.$$

$\endgroup$
  • $\begingroup$ Thank you, that actually helped get me started! $\endgroup$ – Faith Sep 10 '15 at 0:15
0
$\begingroup$

As the others have suggested, we need to use the anti-power rule:

$$\int{x^a}dx=\frac{1}{a+1}x^{a+1}$$

Also as the comment has suggested, we may use $x=(x+a)-a$ to substitute the first $x$:

$$\int_{0}^{1}x(x+a)^{\frac{1}{p}}dx = \int_{0}^{1}[(x+a)-a](x+a)^{\frac{1}{p}}dx$$

This works because $(x+a)-a = x+a-a = x$. Let us continue with the expansion:

$$\int_{0}^{1}[(x+a)(x+a)^{\frac{1}{p}} - a(x+a)^{\frac{1}{p}}]dx$$ $$=\int_{0}^{1}(x+a)^{\frac{1}{p}+1}dx - a\int_{0}^{1}(x+a)^{\frac{1}{p}}dx$$

Now we use the anti-power rule to continue: ($\frac{1}{p}+1=\frac{1+p}{p}$)

$$\int_{0}^{1}(x+a)^{\frac{1+p}{p}}dx - a\int_{0}^{1}(x+a)^{\frac{1}{p}}dx$$ $$=[(1+\frac{1+p}{p})(x+a)^{1+\frac{1+p}{p}} - a((1+\frac{1}{p})(x+a)^{1+\frac{1}{p}}]|_{0}^{1}$$ $$=\frac{1+2p}{p}(1+a)^{\frac{1+2p}{p}}-\frac{1+2p}{p}(a)^{\frac{1+2p}{p}}-a[\frac{p+1}{p}(1+a)^{\frac{p+1}{p}}-\frac{p+1}{p}(a)^{\frac{p+1}{p}}]$$ $$=\left(\frac{1}{p}+2\right)(1+a)^{\frac{1}{p}}(1+a)^2 - \left(\frac{1}{p}+1\right)(1+a)^{\frac{1}{p}}(1+a)a - \left(\frac{1}{p}+2\right)a^{\frac{1}{p}}a^2 + \left(\frac{1}{p}+1\right)a^{\frac{1}{p}}a^2$$ $$=(1+a)^{\frac{1}{p}}(1+a)^2 \left( \left(\frac{1}{p}+2\right) - \left(\frac{a}{1+a}\right) \right) - a^{\frac{1}{p}}a^2$$ $$=(1+a)^{\frac{1}{p}}(1+a)\left( \frac{1+a+2p+ap}{p} \right) - a^{\frac{1}{p}}a^2$$

$\endgroup$
  • $\begingroup$ That's awesome! Thank you for providing such a detailed solution, it was very helpful. $\endgroup$ – Faith Sep 10 '15 at 0:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.