calculus: prove int[0,a][f(x)]dx = int[0,a][f(a-x)]dx if....

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khauna

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Sorry for asking so many questions, but thats what I get for leaving these projects to do at the end... Our teacher doesn't really give us due dates for a lot of things so we just have to finish it and turn it in, which makes procrastination so much easier :roll: I think im just brain dead, after taking integral of squareroot [tanx]....took 3 classmates and I one afternoon on thursday and all this morning...just horrible integral. Anywho, heres another problem, this should be my last one for a little while, until differential equations starts :shock:

This problem stems back to a question i had before for:
integrate: sinx / sinx + cos x (bounds from 0 to pi/2) which we found to be pi/4

that was part of the problem, the rest i still havent been able to figure out.

he first says how the integral sinx / sinx + cosx and sinx^10 / sinx^10 + cosx^10 may be difficult integrals to handle, yet sinx^2 / sinx^2 + cosx^2 is an easy integral. Our teacher wants us to find a "special" technique for integrating
sinx^n / sinx^n + cosx^n

let f be a continuous function over a closed interval [0, a]:

integrate [f(x)dx (from 0 to a)] = integrate [f(a - x)dx (from 0 to a)]

- he wants us to prove that this is true and use that result to somehow evaluate: integral sinx^n / sinx^n + cosx^n

i haven't been able to prove that the integral is true much less evaluate the sin / sin + cos stuff...
 
Just to clarify, he wants you to prove that:

\(\displaystyle \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\), assuming f is continuous over the closed interval [a,b]

Then, use this to come up with a general form solution for:

\(\displaystyle \int\frac{sin^{n}(x)}{sin^{n}(x)+cos^{n}(x)}dx\)

To evaluate we will have to have some limits of integration. Is it an indefinite integral or does it have limits like

\(\displaystyle 0 \;\ to \;\ \frac{\pi}{2}\)

Is that the entire problem statement?.

He's not asking for much, huh?. :)
 
\(\displaystyle \int_{0}^{a}f(x)dx=\int_{0}^{a}f(x-a)dx\), assuming f is continuous over the closed interval [a,b]
This is not true. Example: \(\displaystyle f(x)=x^3\: a=2\)

Do you mean \(\displaystyle \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\)??
If so, try u-substitution, \(\displaystyle u = a - x\).
 
yea thats what i mean, the last one you put ,

integral from 0 to a of f(x) dx = integral from 0 to a of f(a - x) dx
 
Good. Do it with u-substitution: \(\displaystyle u=a-x\).
 
integral f(x),x,0,a = F(a)-F(0), FTCI

integral f(a-x),x,0,a. Let u = a-x, then du = -dx, hence -integral f(u),u,a,0 = integral f(u),u,0,a = F(a)-F(0), FTCI.

Hence both integrals are equal. Second part, I'm working on.
 
Let's try this.

Let \(\displaystyle I=\int_{0}^{a}\frac{f(x)}{f(x)+f(a-x)}dx\)

Let \(\displaystyle u=a-x, \;\ x=a-u, \;\ -du=dx\)

\(\displaystyle I=-\int_{a}^{0}\frac{f(a-u)}{f(a-u)+f(u)}du\)

\(\displaystyle =\int_{0}^{a}\frac{f(a-u)+f(u)-f(u)}{f(a-u)+f(u)}du\)

\(\displaystyle =\int_{0}^{a}du-\int_{0}^{a}\frac{f(u)}{f(a-u)+f(u)}du\)

\(\displaystyle I=a-I\), so \(\displaystyle 2I=a, \;\ I=\frac{a}{2}\)

Now, we could use this to evaluate \(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sin(x)}{sin(x)+cos(x)}dx\)

by noting that \(\displaystyle sin(x)=cos(\frac{\pi}{2}-x)\)

by just plugging in \(\displaystyle a=\frac{\pi}{2}\) and getting \(\displaystyle I=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}\)

To show,\(\displaystyle \int_{0}^{\frac{\pi}{2}}sin^{n}(x)dx=\int_{0}^{\frac{\pi}{2}}cos^{n}(x)dx\)

\(\displaystyle sin(x)=cos(\frac{\pi}{2}-x)\):

\(\displaystyle \int_{0}^{\frac{\pi}{2}}cos^{n}(\frac{\pi}{2}-x)dx=\)

Let \(\displaystyle u=\frac{\pi}{2}-x\)

\(\displaystyle -\int_{\frac{\pi}{2}}^{0}cos^{n}(u)du\)

\(\displaystyle =\int_{0}^{\frac{\pi}{2}}cos^{n}(u)du\)

\(\displaystyle =\int_{0}^{\frac{\pi}{2}}cos^{n}(x)dx, \;\ \text{replace u by x}\)

And we can show that \(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sin^{n}(x)}{cos^{n}(x)+sin^{n}(x)}dx=\frac{\pi}{4}\), for positive n.
 
integral of [sin(x)]^n/{[sin(x)]^n + [cos(x)]^n},x,0,Pi/2 = integral of [sin((Pi/2)-x)]^n/{[sin((Pi/2)-x)]^n + [cos((Pi/2)-x)]^n},x,0,Pi/2 = [cos(x)]^n/{[sin(x)]^n + [cos(x)]^n},x,0,Pi/2.

Hence First integral = last integral implies [sin(x)]^n = [cos(x)]^n, x = Pi/4

Therefore integral of [1/2^(n/2)]/{1/2^(n/2) + 1/2^(n/2)},x,0,Pi/2 = integral (1/2),x,0,Pi/2 = Pi/4

Ergo integral of [sin(x)]^n/{[sin(x)]^n + [cos(x)]^n},x,0,Pi/2 = Pi/4, n any Real.

Pretty cute.
 
\(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sin^{n}(x)}{sin^{n}(x) + cos^{n}(x)}dx\)

\(\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{sin^{n}((\frac{\pi}{2})-x)}{sin^{n}((\frac{\pi}{2})-x) + cos^{n}((\frac{\pi}{2})-x)}dx\)

\(\displaystyle =\int_{0}^{\frac{\pi}{2}} \frac{cos^{n}(x)}{sin^{n}(x) + cos^{n}(x)}dx\)

Hence First integral = last integral implies \(\displaystyle sin^{n}(x) = cos^{n}(x), \;\ x = \frac{\pi}{4}\)

Therefore, \(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{(1/2)^{n/2}}{(1/2)^{n/2} + (1/2)^{n/2}}dx= \int_{0}^{\frac{\pi}{2}} (1/2)dx = \frac{\pi}{4}\)

Ergo, \(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{sin^{n}(x)}{sin^{n}(x) + cos^{n}(x)}dx= \frac{\pi}{4}, \;\ \forall{n}\in\mathbb{R}\)

Pretty cute.
Click to expand...

Yes, that is very nice. But folks could appreciate it more if it were in LaTex. Much easier to read and easier to appreciate your nice proof.
 
khauna said:
how do u use LaTex?
Click to expand...
One way to learn how to use LaTeX would be to read articles online (found through Google, for example) or the relevant articles listed in the "Forum Help" pull-down menu at the very top of every page on this forum.

Eliz.
 
I type the code myself. But you can also use software called mathtype to do it.

Click on 'quote' in the upper right hand corner of my posts to see what I typed to make it display that way. It is more typing, but worth it so one can see it better rather than try to muddle through keyboard font.

For instance,
Code: 
\int_{a}^{b}f(x)dx[\code]

will display this:

[tex]\int_{a}^{b}f(x)dx[/tex]

instead of INT[f(x)]dx, x=a..b

Except use [tex] instead of [code]

I done that so it would display the code and you could see it.
 
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