Question

Find the definite integral

Answer 1

\[\int\limits x^2 e^{-x^3}dx\]

Answer 2

ok lets dot his integration by parts.

u = x^2
1/2du = xdx

dv = \[e^{-x^3}dx\]

does v = \[e^{\frac{-x^4}{4}}dx\]

Answer 3

http://www.wolframalpha.com/input/?i=integration+of+e%5E%28-x%5E3%29

I don't like Wolfram's answer....

Answer 4

the other way I was thinking was to do a u sub.

Answer 5

u = -x^3
du = -3x^2

inegration of e^u

Answer 6

set your u-sub to be -x^3

Answer 7

that's better..

Answer 8

:D

Answer 9

I would use an u-Substitution to tackle this problem. Let:
\[u=-x^3\]
then
\[du/dx = -3x^2 \rightarrow (-1/3)(du/dx)=x^2\]
such that:
\[(-1/3)\int\limits_{}^{}e^{u}(du/dx)*dx = (-1/3)\int\limits_{}^{}e^e du\]

Answer 10

\[\frac{1}{3} \int\limits e^u du\]

Answer 11

\[\frac{1}{3} e^u + c\]
\[\frac{1}{3} e^{-x^3}\]

Answer 12

= v?

Answer 13

I believe there is a minus sign missing.

Answer 14

oh yeah I suck LOL

Answer 15

I always forget that ittle stuff :P

Answer 16

no you did perfect, happens to me all the time, but good you noticed the thing about the u-substitution.

Answer 17

Yeah I figured it was just doing it normally, or most likely a u-sub within :)

Answer 18

so now we do \[u*v - \int\limits vdu\]

Answer 19

\[x^2 * -\frac{1}{3}e^{-x^3} + \frac{1}{3} * \frac{1}{2} \int\limits e^{-x^3}xdx\]

Answer 20

now it seems like I have to do integration by parts once again huh?

Answer 21

this is super ugly :P

Answer 22

oh wait this is a definite integral too LOL from 0-3 woops :p

Answer 23

lets simplify this bad boy.

Answer 24

\[-\frac{x^2}{3}e^{-x^3} + \frac{1}{6} \int\limits e^{-x^3}xdx\]

Answer 25

For the last integral you have to use integration by parts, since the derivative of the exponent cannot be lineary transformed into the x.

Answer 26

yup so I was correct

u = x
du = dx

v = \[-\frac{1}{3}e^{-x^3}\]

Answer 27

\[-\frac{x^2}{3}e^{-x^3} + \frac{1}{6} -\frac{x}{3}e^{-x^3} + \frac{1}{3} \int\limits e^{-x^3} dx\]

Answer 28

\[-\frac{x^2}{3}e^{-x^3} + \frac{1}{6} -\frac{x}{3}e^{-x^3} + \frac{1}{9} e^{-x^3} +\]

Answer 29

c

Answer 30

Do you have the complete problem somewhere? Above you only asked for the integral:
\[\int\limits_{}^{}x^2e^{-x^3}dx\]

Answer 31

yes, we has to do integration by parts... twice?

Answer 32

no, not for this problem, the one I just reposted?

Answer 33

yes.

Answer 34

no, after one u-substitution you are done.

Answer 35

yes I just realized that when you asked hmm.t LOL

Answer 36

Ugh...

Answer 37

BIg headache FTL.

Answer 38

I was a bit confused and thought you continue on another problem now (-: Anyhow, the answer you have give above is already 100% correct.

Answer 39

so u = -x^3
du = -3x^2

\[\int\limits e^u du\]

Answer 40

\[ -\frac{1}{3}e^{-x^3} \]

Answer 41

I suck LOL

Answer 42

exactly, that is the answer to the problem above.

Answer 43

this is a definite integral from 0-3 I forgot that at the top

Answer 44

\[ -\frac{1}{3}e^{-3^3} +\frac{1}{3}e^{-1^3} \]

Answer 45

yes now you can evaluate that from 0 to 3. F(a)-F(b).

Answer 46

\[ -\frac{1}{3}e^{-9} +\frac{1}{3}e^{-1} \]

Answer 47

man tons of wasted time on this problem thinking it was integration by parts LOL

Answer 48

wait, you said, zero. so it should be only +1/3 at the second expression

Answer 49

yea 0 woops :P again I suck.

Answer 50

e^0 is 1 :P

Answer 51

\[ -\frac{1}{3}e^{-9} +\frac{1}{3}\]

Answer 52

hehe, exactly. The longer I work on a problem, they more complicated I make them myself too.

Answer 53

yes that is the answer.

Answer 54

Thanks :D

Answer 55

very welcome