Question

How do you integrate `e^(x^2)` from 0 to 1?

Answer 1

`e^1`

Explanation:

`int_0^1e^(x^2)`
`[e^(x^2)]_0^1`
`[e^(1^2)-e^(0^2)]`
=`e^1`

Answer 2

We can't find an exact value for `int_0^1e^(x^2)dx` because `int_1^x e^(x^2)dx` cannot be described in terms of elementary functions.

So the best we can do is use a Maclaurin series approximation.

Recall that

`e^x = sum_(n = 0)^oo x^n/(n!) = 1 + x + x^2/(2!) + x^3/(3!)`

Thus

`e^(x^2)= sum_(n = 0)^oo x^(2n)/(n!) = 1 + x^2 + (x^4)/(2!) + (x^6)/(3!)`

Now you integrate

`int_0^1 e^(x^2)dx = [x + 1/3x^3 + 1/(5(2!))x^5 + 1/(7(3!))x^7]_0^1`

`int_0^1 e^(x^2)dx = 1/42 + 1/10 + 1/3 + 1 ~~1.457`

A calculator should give an approximation of `1.463`, so our answer isn't too terrible. Increasing the number of terms of the maclaurin series in the application will make the approximation more precise.

Hopefully this helps!