Question

Use an infinite series to evaluate the definite integral of x^2e^(-x^2) from 0 to 1.

The above problem looks like something I've done before, but I'm having trouble finding it in my notes. Perhaps someone could give me a clue as to how to get started?

Answer 1

\[\int\limits_{0}^{1} x^2e^{-x^2}\]

For those having trouble understanding the question as I typed it. I'm supposed to use an infinite series to evaluate it.

Answer 2

Recall the infinite series definition of e^x :
\[\large e^x = \sum \limits_{n=0}^{\infty }\dfrac{x^n}{n!}\]

Answer 3

plugin x = -x^2,
multiply both sides by x^2,
then take the integral

Answer 4

Why do I multiply both sides by x^2?

Answer 5

look at the integral you're trying to evaluate

Answer 6

the idea is to cookup that integral from this series representation of e^x

Answer 7

Ah. Okay. So when I integrate both sides, do I still include sigma in the notation, or am I only taking the integral of x^2(-x^2)^n/n! on the right side?

Answer 8

when the series converges :
\[\large \int \sum = \sum \int \]

Answer 9

\[\large e^x = \sum \limits_{n=0}^{\infty }\dfrac{x^n}{n!}\]

\[\large e^{-x^2} = \sum \limits_{n=0}^{\infty }\dfrac{(-x^2)^n}{n!} = \sum \limits_{n=0}^{\infty }\dfrac{(-1)^nx^{2n}}{n!} \]

\[\large x^2e^{-x^2} = \sum \limits_{n=0}^{\infty }\dfrac{(-1)^nx^{2n+2}}{n!} \]

Answer 10

integrate both sides

Answer 11

\[\large \int_0^1x^2e^{-x^2}dx = \int_0^1\sum \limits_{n=0}^{\infty }\dfrac{(-1)^nx^{2n+2}}{n!} dx\]

Answer 12

\[\large = \sum \limits_{n=0}^{\infty } \int_0^1 \dfrac{(-1)^nx^{2n+2}}{n!} dx \]

Answer 13

\[\large = \sum \limits_{n=0}^{\infty } \dfrac{(-1)^n}{n!} \int_0^1 x^{2n+2} dx \]

Answer 14

whats the integral of \(\large \int_0^1 x^{2n+2} dx \) ?

Answer 15

\[\int\limits_{0}^{1} x^{2n+2} = \frac{ x^{2n+3} }{ 2n+3 } evaluated from 0 \to 1\]

Answer 16

yes evaluate the bounds

Answer 17

After evaluating the bounds, you get \[\frac{ (1)^{2n+3} }{ 2n+3 }\]

n doesn't have a value with regards to the integral, so I assume you multiply that result with the infinite series to get:

\[\sum_{0}^{\infty} \frac{ (-1)^n(1)^{2n+3} }{ n!(2n+3) }\]

Answer 18

that looks good ^^ but hey (1)^anything = 1 right ?

Answer 19

Right! So, after that it would simplify to: \[\sum_{0}^{\infty} \frac{ (-1)^n }{ n!(2n+3) }\]

Answer 20

Perfect !

Answer 21

And I'm guessing all I need now to finish the evaluation is to find the sum of the infinite series and I'm done?

Answer 22

You're done! leave it as infinite series...

Answer 23

Cool! Thanks.

Answer 24

yw!