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

Apr 10, 2018

${e}^{1}$

#### Explanation:

${\int}_{0}^{1} {e}^{{x}^{2}}$
${\left[{e}^{{x}^{2}}\right]}_{0}^{1}$
$\left[{e}^{{1}^{2}} - {e}^{{0}^{2}}\right]$
=${e}^{1}$

Apr 10, 2018

We can't find an exact value for ${\int}_{0}^{1} {e}^{{x}^{2}} \mathrm{dx}$ because ${\int}_{1}^{x} {e}^{{x}^{2}} \mathrm{dx}$ 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

${\int}_{0}^{1} {e}^{{x}^{2}} \mathrm{dx} = \frac{1}{42} + \frac{1}{10} + \frac{1}{3} + 1 \approx 1.457$

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!