Question

integral (x^2 e^-x^2) dx

Answer 1

Integrate by parts once. The resulting integral will be a simple substitution.

Answer 2

That's not true.

Answer 3

Ah, missed the minus sign...

Answer 4

is u=x^2 and dv=?

Answer 5

It's not the minus sign, is the fact that its
\[ e^{x^2} \]
it's called a Gaussian integral.

Answer 6

@blinn13 disregard my suggestion

Answer 7

oops, sorry,
\[ e^{-x^2} \]

Answer 8

okay

Answer 9

the integral is not in closed form.

Answer 10

*cannot be expressed

Answer 11

What are the limits of integration?

Answer 12

0 to 4....im trying to find a centroid

Answer 13

hmm ... i would use calculator

Answer 14

i have already found the number i need to show work though :/

Answer 15

numerical methods?

Answer 16

Well then there's a mistake somewhere because it can't be done by hand.

Answer 17

i see...is (e^-x)^2 equal to e^-2x or e^-x^2 then?

Answer 18

e^(-2x) of course

Answer 19

thats where my problem was then...thank you

Answer 20

then it just becomes u-sub.

Answer 21

yes yes

Answer 22

Ah. Yep, there we go.

Answer 23

or by parts

Answer 24

by parts would be better

Answer 25

so u=x^2, du=2x dx, dv=e^-2x dx, v=-e^-2x/2..right?

Answer 26

So by parts was right, eh? hehe

Answer 27

how do you use IE by parts?
here we use
\[ \int u v dx = u \int v - \int (du/dx \int v ) \]

Answer 28

i have never seen that before

Answer 29

u = x^2
v = e^(-2x)

Answer 30

would i be able to do it with normal integration by parts?

Answer 31

yes .. twice

Answer 32

ok ill try that...my previous values for u and dv are correct?

Answer 33

i know that type of form exist ... but i don't use it often.

Answer 34

\[ \begin{align*}
\int x^2 e^{-2x}dx dx &= x^2 \int e^{-2x}dx - \int \left( \frac{dx^2}{dx} \int e^{-2x}dx\right )dx \\
&= x^2 \frac{e^{-2x}}{-2 } - \int 2x \frac{e^{-2x}}{-2 }dx\\
&= - \frac{x^2 e^{-2x}}{2 } + \int x e^{-2x}dx \\
&= - \frac{x^2 e^{-2x}}{2 } + x \int e^{-2x}dx - \int \left( \frac{dx}{dx}\int e^{-2x} dx\right )dx\\
\end{align*} \]

Answer 35

we have not gotten that far but i think it is coming up soon

Answer 36

this is integration by parts

Answer 37

i learned it a different way then : \[uv -\int\limits vdu\]

Answer 38

yes yes i know that ... integration by parts is just opposite of product rule in differentiation

Answer 39

i know that

Answer 40

thank you for your help!

Answer 41

\[ d(uv) = u dv + v du \\
\int d(vu) = \int u dv + \int v du \\
uv = \int u dv + \int v du \\
\int v du = uv - \int u dv
\]
put v=g(x), u = \( \int f(x) dx \) you get
\[
\int g d\left( \int f \right ) = g \int f dx - \int \left(\frac{dg}{dx} \int f dx\right ) dx \]
I think is is easier to use.

Answer 42

\[
\int f g dx = g \int f dx - \int \left(\frac{dg}{dx} \int f dx\right ) dx \]