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I am pretty new to calculus and would like a nudge in the right direction in order to complete this question properly (Maybe also correct any misrepresentations I have about integration)

So the question is this:

$$\int_{0}^{1}x\sqrt{1-\sqrt{x}}\:dx$$

My attempt at a solution:

let $$u=\sqrt{x}$$ $$u^2={x}$$

thus $$\frac{du}{dx}=\frac{1}{2}x^{-1/2}$$ $$dx=x^{1/2}2du=2udu$$

Now, $$\int_{a}^{b}u^{2}\sqrt{1-u}\:2udu\:$$

$$=2\int_{a}^{b}u^{2}\sqrt{1-u}\:udu$$

$$=2(\frac{1}{3}u^{3}*\frac{2}{3}({1-u})^{3/2}*\frac{1}{2}u^{2}*\frac{1}{2}u^{2})$$

I am able to substitute $\sqrt{x}$ for u, and solve the integral, but i don't believe this gives me the correct solution.

Can anyone point me towards a mistake, mis-conceptualization or different substitution that would allow me to solve this question?

The question is tagged as dual substitution, does that mean i need to substitute two factors, or that multiple substitutions will work for this integral?

Thanks

P.S. Feel free to clean up my LaTex

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    $\begingroup$ Where did that last line "$=2(\frac{1}{3}u^{3}*\frac{2}{3}({1-u})^{3/2}*\frac{1}{2}u^{2}*\frac{1}{2}u^{2})$" come from? $\endgroup$ – user137731 Jan 12 '16 at 19:04
  • $\begingroup$ The method looks OK to me, for the last step you are going to have to integrate by parts. I am not sure the final answer given there is correct. Also, why have you left your bounds near the end as $a,b$? The substitution $u = 1-\sqrt{x}$ also works. $\endgroup$ – fosho Jan 12 '16 at 19:06
  • $\begingroup$ Bye_world: I am not sure if i integrated properly considering i used the chain rule to integrate $(1-u)^{1/2}$ $\endgroup$ – helpmeh Jan 12 '16 at 19:09
  • $\begingroup$ Workaholic, ill try that and post my attempt $\endgroup$ – helpmeh Jan 12 '16 at 19:10
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    $\begingroup$ @helpmeh The chain rule states that $f(g(x))'=f'(g(x))\cdot g'(x)$, how did you use it to find the antiderivative? $\endgroup$ – Workaholic Jan 12 '16 at 19:11
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In my previous answer I ended where I assumed you could take over. I guess that was a false assumption. In this answer I'll try to explain each step.

Start with the integral $$\int_{x=0}^{x=1} x\sqrt{1-\sqrt{x}}\ dx$$

Here I'd be a bit more ambitious than you. I know that anything other than simply a variable under a square root is going to cause me trouble so I'm going to try to substitute everything under the outer square root.

Let $u = 1-\sqrt{x}$. This implies that $$du = \frac{d(1-\sqrt{x})}{dx}\ dx = -\frac{1}{2\sqrt{x}}\ dx = -\frac 1{2(1-u)}\ dx \\ \implies -2(1-u)\ du = dx$$

We also will need to substitute for $x$ so notice that $u=1-\sqrt{x} \implies x=(1-u)^2$.

Now we make this substitution. Notice how I evaluate the bounds:

$$\begin{align}\int_{x=0}^{x=1} (1-u)^2\sqrt{u}\big(-2(1-u)\ du\big) &= -2\int_{u=1-\sqrt{0}}^{u=1-\sqrt{1}} (1-u)^3\sqrt{u}\ du \\ &= -2\int_{u=1}^{u=0} (1-u)^3u^{1/2}\ du \\ &= 2\int_{u=0}^{u=1} (1-u)^3u^{1/2}\ du\end{align}$$

From here we should just simplify $(1-u)^3u^{1/2}$. $$(1-u)^3u^{1/2} = (1-3u+3u^2-u^3)u^{1/2} = u^{1/2}-3u^{3/2}+3u^{5/2}-u^{7/2}$$

Now we can use this and the fact that $\int_a^b (\beta f(x) + \gamma g(x))\ dx = \beta\int_a^b f(x)\ dx + \gamma\int_a^b g(x)\ dx$ to get our integral into a form we know how to evaluate.

$$\begin{align}2\int_{u=0}^{u=1} (1-u)^3u^{1/2}\ du &= 2\int_0^1 (u^{1/2}-3u^{3/2}+3u^{5/2}-u^{7/2})\ du \\ &= 2\left(\int_0^1 u^{1/2}\ du -3\int_0^1 u^{3/2}\ du + 3\int_0^1 u^{5/2}\ du -\int_0^1 u^{7/2}\ du\right) \\ &= 2\int_0^1 u^{1/2}\ du -6\int_0^1 u^{3/2}\ du + 6\int_0^1 u^{5/2}\ du - 2\int_0^1 u^{7/2}\ du\end{align}$$

At this point we remember that the antiderivative of $x^k$ is $\frac 1{k+1}x^{k+1}$ and get the answer:

$$2\int_0^1 u^{1/2}\ du -6\int_0^1 u^{3/2}\ du + 6\int_0^1 u^{5/2}\ du - 2\int_0^1 u^{7/2}\ du\ = \left.\big(2\frac 23u^{3/2} -6\frac25u^{5/2} +6\frac27u^{7/2}-2\frac29u^{9/2}\big)\right|_{\ 0}^{\ 1} \\ = \frac 43-\frac{12}5+\frac{12}7-\frac 49$$

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    $\begingroup$ thanks for your help you are a good person :) $\endgroup$ – helpmeh Jan 12 '16 at 21:16
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At the line $$=2\int_{0}^{1}u^{2}\sqrt{1-u}\:udu = 2\int_0^1 u^3\sqrt{1-u}\ du$$

(why did I set the bounds $=0$ and $1$?) you could do integration by parts, but since you say you're a "beginner", why not just use another substitution: Let $w=1-u$, then $dw=-du$.

$$=-2\int_1^0 (1-w)^3\sqrt{w}\ dw = 2\int_0^1 (1-w)^3\sqrt{w}\ dw$$ Then if you expand this out (meaning $(1-w)^3\sqrt{w} = (1 -3w + \cdots)\sqrt{w} = w^{1/2}-3w^{3/2}+\cdots$) you'll just be left with several integrals of the type $\int aw^k\ dw$ which hopefully you know how to evaluate. Good luck. :)

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  • $\begingroup$ this is helpful, AFAIK the bounds do not need to be set until I am ready to solve the integer. However i have never seen the notation dw, I am assuming it is simply a second substitution, I am going to have some food and come back to this question with a vengeance $\endgroup$ – helpmeh Jan 12 '16 at 19:15
  • $\begingroup$ The bounds I obtained by simply setting $x=0$ and $1$ in the change-of-variables equation $u=\sqrt{x}$. Later I found the bounds in the integration wrt $w$ by setting $u=0$ and $1$ in the change-of-variables equation $w=1-u$. As for what $dw$ is -- it's the same thing as $du$ and $dx$, I just had to choose a new name for the variable of integration because $x$ and $u$ were already taken. $\endgroup$ – user137731 Jan 12 '16 at 19:19
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attempt numero dos, (thanks to @bye_world) (third time maybe charm)

from $$=2\int_{a}^{b}u^{2}\sqrt{1-u}\:udu$$

$$=2\int_{a}^{b}u^{3}\sqrt{1-u}\:du$$

let $$w=1-u\:$$ $$u=1-w\:$$ and $$\frac{dw}{du}=-1$$ $${-dw}={du}$$

so, $$=2\int_{b}^{a}({1-w})^{3}\sqrt{w}\:dw$$ $$=2\int_{b}^{a}(-w^{7/2}+3w^{5/2}-3w^{3/2}+w^{1/2})\:dw$$

Okay, I think I get what you guys are trying very hard to get through to me, $\int{f(x)+g(x)+h(x)}$ can be split up into $\int{f(x)}+\int{g(x)}+\int{h(x)}$ but $\int{f(x)*g(x)}$ does not equal $\int{f(x)}*\int{g(x)}$

Anyways so $$=2[(-\frac{2}{9}(w)^{9/2}+\frac{6}{7}(w)^{7/2}-\frac{6}{5}(w)^{5/2}+\frac{2}{3}(w)^{3/2})]^0_1$$

amirite?

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  • $\begingroup$ edited, please explain "you can't antidifferentiate an integrand factor by factor." In a way I can understand (understood) :p... Thanks for all the help by the way $\endgroup$ – helpmeh Jan 12 '16 at 20:05
  • $\begingroup$ Given two functions $f(x)$ and $g(x)$, for which you know the antiderivatives $\int f(x)\ dx = h(x)$ and $\int g(x)\ dx = i(x)$, you cannot say that $\int f(x)g(x)\ dx = h(x)i(x)$. Integration, sadly, just isn't that simple. That's why in my answer I said you should expand out $(1-w)^3\sqrt{w}$ instead of just evaluating it directly at the stage $\int (1-w)^3\sqrt{w}\ dw$. $\endgroup$ – user137731 Jan 12 '16 at 20:08
  • $\begingroup$ That (your latest edit) doesn't work either. Reread the last paragraph of my answer. $\endgroup$ – user137731 Jan 12 '16 at 20:19
  • $\begingroup$ edited, check it out, or are you saying i should expand $(1-w)^{3}=-w^{3}+3w^{2}-3w+1$ $\endgroup$ – helpmeh Jan 12 '16 at 20:22
  • $\begingroup$ @helpmeh $\int f(x) \cdot g(x) dx \neq \int f(x) dx \cdot \int g(x) dx$ This is what bye is saying above. $\endgroup$ – randomgirl Jan 12 '16 at 20:25

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