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

2 Answers
Shubh
Apr 10, 2018

#e^1#

Explanation:

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

Noah G
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!

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