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

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

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Answer 1 · 2018-04-10T15:38:53

$e^{1}$ $e^1$

Explanation:

$\int_{0}^{1} e^{x^{2}}$ $int_0^1e^(x^2)$
${[e^{x^{2}}]}_{0}^{x^{2}}$ $[e^(x^2)]_0^1$
$[e^{1^{2}} - e^{0^{2}}]$ $[e^(1^2)-e^(0^2)]$
= $e^{1}$ $e^1$

Answer 2 · 2018-04-10T15:47:55

We can't find an exact value for $\int_{0}^{1} e^{x^{2}} d x$ $int_0^1e^(x^2)dx$ because $\int_{1}^{x} e^{x^{2}} d x$ $int_1^x e^(x^2)dx$ cannot be described in terms of elementary functions.

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

Recall that

$e^{x} = \infty \sum n = 0 \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!}$ $e^x = sum_(n = 0)^oo x^n/(n!) = 1 + x + x^2/(2!) + x^3/(3!)$

Thus

$e^{x^{2}} = \infty \sum n = 0 \frac{x^{2 n}}{n!} = 1 + x^{2} + \frac{x^{4}}{2!} + \frac{x^{6}}{3!}$ $e^(x^2)= sum_(n = 0)^oo x^(2n)/(n!) = 1 + x^2 + (x^4)/(2!) + (x^6)/(3!)$

Now you integrate

$\int_{0}^{1} e^{x^{2}} d x = {[x + \frac{1}{3} x^{3} + \frac{1}{5 (2!)} x^{5} + \frac{1}{7 (3!)} x^{7}]}_{0}^{3}$ $int_0^1 e^(x^2)dx = [x + 1/3x^3 + 1/(5(2!))x^5 + 1/(7(3!))x^7]_0^1$

$\int_{0}^{1} e^{x^{2}} d x = \frac{1}{42} + \frac{1}{10} + \frac{1}{3} + 1 \approx 1.457$ $int_0^1 e^(x^2)dx = 1/42 + 1/10 + 1/3 + 1 ~~1.457$

A calculator should give an approximation of $1.463$ $1.463$ , 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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