Approximation methods that can be applied

1. Sep 11, 2004

hedlund

How do I find integrals like
$$\int_{a}^{b} \left( x^2 + 1 \right)^2 \ dx$$. This one is easy, since I can just turn it into $$\int_{a}^{b} \left( x^4 + 2x^2 + 1 \right) \ dx$$. But what if it would say $$\int_{a}^{b} \left( x^2 + 1 \right)^{40} \ dx$$? What technique should I use?

2. Sep 11, 2004

devious_

There are certain approximation methods that can be applied.

3. Sep 11, 2004

matt grime

Or substituion methods to obtain a (possibly hyperbolic) trig integral. Often you'll end up with trying to integrate something that we can write as:

I(n) = int f(x)dx

where f( x) is some function of x, and the n indicates some exponent or coefficient (the 40 in your case), then doing something like integration by parts you'll end up with

I(n-1) + something= I(n)

and you get an iteration going that lets you figure out how to solve the integral.

4. Sep 11, 2004

matt grime

I think you need to fix that latex.

5. Sep 11, 2004

Muzza

Yeah, I tried to (and then I deleted my post before I noticed that you had replied to it. Doh). LaTeX doesn't seem to like the combination of an integral sign, sum sign and a binomial coefficient... Let's try again:

$$\int (x^2 + 1)^{40} dx = \int \sum^{40}_{k=0} C^{40}_{k} x^{2k} dx = \left( \sum^{40}_{k=0} \frac{C^{40}_{k}}{2k + 1} x^{2k + 1} \right) + D.$$

Actually calculating those binomial coefficients is going to suck though :P

Last edited: Sep 11, 2004
6. Sep 11, 2004

Theelectricchild

Leaving it as a series solution is absolutely fine--- consider

$$\int e^{x^2} dx$$

There is no substitution you can use, but you can use the general series form of e^x and plug in x^2 and then integrate the sum as a whole, and leave you answer as that.

7. Sep 11, 2004

JonF

this maybe silly, but wouldn't a trig sub then a integral table work for something like that?

8. Sep 11, 2004

Dr Transport

try the trig substitution for the original problem, let
$$x = \tan(y)$$, then the portion in the parentheses becomes $$\sec(y)^2$$ from there you can make the substitutions and complete the integral. As for the $$e^{x^{2}}$$ over a finite interval, you have to use an integral table because it can only be done numerically, if the limits are infinite in extent, then the integral is $$\frac{\pi}{2}$$,

9. Sep 11, 2004

Theelectricchild

$$\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$$

10. Sep 12, 2004

HallsofIvy

Staff Emeritus
Yes, but the theelectricchild's point was that there is no "elementary" anti-derivative for $e^{x^2}$. The sad fact is that the great majority (in a very specific sense, "almost all") of nice "elementary" functions, even though they have an anti-derivative, the anti-derivative is not an "elementary" function.

11. Sep 12, 2004

Dr Transport

I forgot the square root.........

12. Sep 12, 2004

Theelectricchild

Hehe its ok, thats one of the first places I have seen a "squareroot" of pi... its an integral you can actually solve using a double integration step.

$$\int_{0}^{\infty} e^{-x^2} dx$$

Convert to DI problem:

$${(\int_{0}^{\infty} e^{-x^2} dx)}^2$$

$$=(\int_{0}^{\infty} e^{-x^2} dx) (\int_{0}^{\infty} e^{-y^2} dy)$$

$$=\int_{0}^{\infty}\int_{0}^{\infty} e^{-x^2}e^{-y^2} dxdy$$

$$=\int_{0}^{\infty}\int_{0}^{\infty} e^{-{(x^2+y^2)}} dxdy$$

From here using a Change of Variables step or noticing the region we are integrating is simply a semicircular arc, we can use polar coordinates:

$$=\int_{0}^{\frac{\pi}{2}}\int_{0}^{R} e^{-r^2}r drd\theta$$

Where we take the limit of this as R --> infinity and the extra r in the integrand is from the Jacobian when switching to polar coordinates (proven in the change of variables theorem)

And NOW we can use a U-substitution to easily solve this iterated double integral:

$${(\int_{0}^{\infty} e^{-x^2} dx)}^2 = \frac{\pi}{4}$$

So $$\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$$

It seems like we can solve quite a few "definite" or "improper" integrals of non elementary functions using multivar. calculus. Does anyone know any other methods? Thanks.

Last edited: Sep 12, 2004
13. Sep 12, 2004

Theelectricchild

I like integrating things.

14. Sep 13, 2004

Divergent13

Where did the R come from again?

15. Sep 15, 2004

Theelectricchild

Well taking an improper integral requires you to take the limit of the evaluated integral as some letter in place of infinity goes to it--- I didnt learn to do limits in TeX yet Ill show it later.