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

2 Answers
Apr 10, 2018

e^1

Explanation:

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

Apr 10, 2018

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!