Substitution Help

**Lord Voldemort** · February 8th 2010, 06:07 PM

I need help again, after pondering this for a while

If a and b are positive numbers, prove that
$\int^1_0 x^a(1-x)^b\ dx = \int^1_0 x^b(1-x)^a\ dx$
I think I should make $u = 1 - x$ and $x = 1 - u$ , then I get:
$-\int^1_0 u^b(1-u)^a\ du$
However, I have been stuck here for a while.
I don't know how to get rid of the negative nor how to substitute in back x without messing it back up.

**tonio** · February 8th 2010, 06:15 PM

Originally Posted by Lord Voldemort

I need help again, after pondering this for a while

If a and b are positive numbers, prove that
$\int^1_0 x^a(1-x)^b\ dx = \int^1_0 x^b(1-x)^a\ dx$
I think I should make $u = 1 - x$ and $x = 1 - u$ , then I get:
$-\int^1_0 u^b(1-u)^a\ du$

This is almost correct, but after the substitution $u=1-x$ , the integral limits get reversed: $x=0\Longrightarrow u=1\,,\,\,x=1\Longrightarrow u=0$ , so all is fine but the last integral's lower limit is 1 and the upper one is 0.

Tonio

However, I have been stuck here for a while.
I don't know how to get rid of the negative nor how to substitute in back x without messing it back up.

.

**Lord Voldemort** · February 8th 2010, 06:26 PM

Thanks! I just realized that after dinner. Funny how taking a break can make things clear. But the first is in terms of x and the second is in terms of u, and u is something in terms of x, so I don't understand how I can say this is complete at this point...

**tonio** · February 8th 2010, 06:32 PM

Originally Posted by Lord Voldemort

Thanks! I just realized that after dinner. Funny how taking a break can make things clear. But the first is in terms of x and the second is in terms of u, and u is something in terms of x, so I don't understand how I can say this is complete at this point...

Instead of writing "u" in the last integral, after substituing, just write "x"...what's the difference ?!

Tonio

**drumist** · February 8th 2010, 06:47 PM

Remember that the variable used in an integral is essentially a "dummy" variable. For example, evaluating a simple integral like this:

$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$

So, what we notice here is that the result doesn't depend on x at all. That whole integral equals 1/3. Therefore, you could change that x to any other variable and get the same result:

$\int_0^1 u^2 \, du = \left[ \frac{u^3}{3} \right]_0^1 = \frac{1}{3}$

or even

$\int_0^1 \phi^2 \, d\phi = \left[ \frac{\phi^3}{3} \right]_0^1 = \frac{1}{3}$

That's why the variable used doesn't matter.

So, in summary, we can say:

$\int_0^1 x^2 \, dx = \int_0^1 u^2 \, du = \int_0^1 \phi^2 \, d\phi = \cdots$

**Lord Voldemort** · February 8th 2010, 06:57 PM

Well, I just thought that because I was trying to equate something then changing up variables between equations wouldn't be such a great idea. But I suppose. Could you verify my answer on this problem as well?

Evaluate $\int^a_0 x\sqrt{a^2-x^2}\ dx$
Let $u = a^2 - x^2$ . Then $du = -2x$
$1/2{\int^{a^2}_0 \sqrt{u}\ dx}$
$\frac{{u^{3/2}}}{3}$ evaluated from 0 to $a^2$ .
$\frac{[{{{a^2 - x^2}}]^{3/2}}}{3}$ evaluated from 0 to $a^2$ .
$\frac{1}{3}a^3[(1-a^2)^{\frac{3}{2}}-1]$

@ drumist: Isn't that equivalent to using x = 1 - x for the substitution? I feel I don't understand this fully

(if you substitute u = x in the end I mean)

**drumist** · February 8th 2010, 08:53 PM

When you changed the variables, you also changed the endpoints of the integral to match. So those endpoints (0 and a^2) are values of u, not values of x.

So you should have done:

$\frac{1}{2} \left[\frac{u^{3/2}}{3}\right]_{u=0}^{u=a^2} = \frac{(a^2)^{3/2}}{6}$

Alternatively, you could have changed the u back to $a^2-x^2$ , but the endpoints need to change to values of x also, like so:

$-\frac{1}{2}\left[\frac{(a^2-x^2)^{3/2}}{3}\right]_{x=0}^{x=a} = \frac{(a^2)^{3/2}}{6}$

As for the second part of your post: Technically there's nothing preventing you from making a substitution from $(1-x) \longrightarrow x$ , but it just gets confusing because you are using x in two contexts, which is why we use u instead.

**drumist** · February 8th 2010, 09:09 PM

I think the confusion comes from a variable being used twice, like this:

$\int_{\textcolor{green}{x}=0}^{\textcolor{green}{x }=5} (\textcolor{green}{x}-1) \, d\textcolor{green}{x} = \int_{\textcolor{red}{x}=-1}^{\textcolor{red}{x}=4} \textcolor{red}{x} \,d\textcolor{red}{x}$

Even though we use x on both sides of the equation, they have nothing to do with one another. If you think about it, the red x's and the green x's don't influence each other at all and operate independently. That's why you can change just one of them, like this:

$\int_{\textcolor{green}{u}=0}^{\textcolor{green}{u }=5} (\textcolor{green}{u}-1) \, d\textcolor{green}{u} = \int_{\textcolor{red}{x}=-1}^{\textcolor{red}{x}=4} \textcolor{red}{x} \,d\textcolor{red}{x}$

And it's still correct.

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