# Solving Int. Exp(-x^2) from 0 to .94 Calculus

• lawtonfogle
In summary: But anyways, wouldn't it be somewhat of a big deal if someone found out how to express that antiderivative in a terms of an elementary function(s)?It definitely would be a major breakthrough, however it's impossible.

## Homework Statement

He happened to give us the problem of finding $$\int _{0} ^{.94}e^{-x^2}dx$$, which I have just been informed by someone who has already taken that class that the teacher does not know how to do it. So, I was wondering, is there a advanced mathematics to find it using integrals, or does one just not exist?

P.S. to understand my level of math, I am in B.C. Calculus.

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this integral is of the form of the "error function" used for the Normal or Gaussian probability density. wikipedia has a nice article:

http://en.wikipedia.org/wiki/Error_function

but, you have to do the scaling correctly. there is no closed form exact solution, but there are series and approximations that get better and better as the number of terms used increases.

there is an exact solution if the upper limit was infinity.

dunno what level in school you are or what class you're taking, but that's a nasty integral to try to blast out without context.

Well, being only a senior in high school...

The teacher was one who might have given that to us and failed to realize what it was (he has done such before).

But on a different note. An exact area is produced by $$\int _{0} ^{a}e^{-x^2}dx$$. Since there is an exact answer, should not there be some way to solve it? Could it just be we do not have a system of mathematics advanced enough to solve it?

it is not about how advance the mathematics is... it has been proved that the integral of e^(-x^2) cannot be expressed in elementary function (sin, cos, +, - exponentials...etc).

However, this integral can easily be expressed in terms of convergent series as rbj said.

and indeed, in many areas of mathematics, most of the time, there just isn't a closed form solution! (as my professor said... if you pick a random function from the functional space, the chance that you will get some nice function is basically zero). and at some instances, it can be proved that no close form solution exists. most of the things we learn in school are solvable simply because they are easier and provide insights in unsolvable problems.

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Just solve the silly thing numerically. Since it's a definite integral, a numeric answer is expected. Give the answer in 4 or 5 sig figs, and call it done. You can use Excel or write your own solution code.

Reasonable advice. Watch out for that loaded comment "...does not know how to do it..." Is there something else in the works such as the erf function or approximation techniques? Students are so quick to disrespect their teachers that it makes me so glad I'm not one. It usually comes down to whether or not you find the teacher handsome enough or whatever else.

Anyway, before I digress, try this:

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i don't know how to do this on my 89, but on an 84, just graph it, then press calc > option 7 > then enter 0 then 0.94

you can't use the fundamental theorem (part 2) cause as it was said, it does not have an antiderivative that can be expressed in elementary functions.

<< inappropriate comment deleted by berkeman >>

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pakmingki said:
<< inappropriate comment deleted by berkeman >>

No, it's a trivial numerical calculation. Please don't paint profs and teachers as "bad guys". Maybe the teacher in this case is trying to find out which students can recognize good candidates for numerical solution versus closed form solutions. I had a couple undergrad classes where that kind of thinking was required.

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i was just playing at the title of the thread of how the teacher is evil (isnt saying that the teacher is evil also painting profs and teachers as bad guys?).
But anyways, wouldn't it be somewhat of a big deal if someone found out how to express that antiderivative in a terms of an elementary function(s)?

pakmingki said:
But anyways, wouldn't it be somewhat of a big deal if someone found out how to express that antiderivative in a terms of an elementary function(s)?

It defnitely would be a major breakthrough, however it's impossible.

Perhaps working out the taylor series, integrating term by term, then put the answer in terms of an infinite series?

Since the upper limit is only 0.94, it should be adequate to, say, use a 3-term Taylor expansion about 0.

i just experienced one similiar. Had to use a calculator.

## 1. What is the purpose of solving Int. Exp(-x^2) from 0 to .94 in calculus?

In calculus, the purpose of solving Int. Exp(-x^2) from 0 to .94 is to find the area under the curve of the function f(x) = e^-x^2. This can be used to solve various problems in physics, engineering, and other fields.

## 2. How does one solve Int. Exp(-x^2) from 0 to .94 in calculus?

To solve Int. Exp(-x^2) from 0 to .94 in calculus, one can use various methods such as the substitution method, integration by parts, or using a table of integrals. It is important to follow the correct steps and use the appropriate method based on the form of the function.

## 3. Can the integral of Exp(-x^2) be solved using basic calculus techniques?

No, the integral of Exp(-x^2) cannot be solved using basic calculus techniques. This is because the function does not have an elementary antiderivative. It can only be solved using more advanced techniques such as the Gaussian integral or numerical methods.

## 4. What is the significance of the limits of integration in solving Int. Exp(-x^2) from 0 to .94?

The limits of integration, in this case 0 and .94, determine the boundaries within which we are finding the area under the curve. This means that the value of the integral will be different depending on the limits chosen. In this case, we are finding the area between the curve and the x-axis from x = 0 to x = .94.

## 5. Can the value of Int. Exp(-x^2) from 0 to .94 be approximated?

Yes, the value of Int. Exp(-x^2) from 0 to .94 can be approximated using numerical methods such as the trapezoidal rule or Simpson's rule. These methods divide the area under the curve into smaller shapes and calculate the approximate area by summing the areas of these shapes. The more shapes used, the more accurate the approximation will be.