# 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.