Question

I'm not sure if the definite integral of these two parts is 0+0=0 or -1/4 + 1/4 = 0.
Question below:

Answer 1

\[\int\limits_{-\infty}^{\infty} x^3e^-x^4 dx\]

Answer 2

\[\int\limits_{-\infty}^{0} x^3e^-x^4 dx = \lim_{a \rightarrow -\infty} \int\limits_{a}^{0}x^3e^-x^4= x^3e^-x^4 \] | a --> 0

Answer 3

\[\int\limits_{0}^{\infty} x^3e^-x^4 dx = \lim_{b \rightarrow \infty} \int\limits_{0}^{b} x^3e^-x^4 = x^3e^-x^4 | 0\rightarrow b\]

Answer 4

is it
\[\int x^3e^{x^4}dx\]?

Answer 5

\[\large \int x^3e^{-x^4}dx\]

Answer 6

if so, it is an odd function

Answer 7

@satellite73 Yes, its the second integral you posted with a negative in front of x^4.

Answer 8

yes, the definite integral equals 0... but u seem to be asking a different question...
can u elaborate a bit ha ? :)

Answer 9

& thank you.

Answer 10

@ganeshie8 * Oh, okay aha. Well my question was the way I got the definite integral as 0. Like the two ways I posted it. Are each integral, when done separately 0 + 0 which equals 0? or what my classmate got, which was -1/4 + 1/4 = 0?

Answer 11

\( \large \mathbb { \int\limits_{-\infty}^{0} x^3e^{-x^4} dx = \lim \limits_{a \rightarrow -\infty} \int\limits_{a}^{0}x^3e^{-x^4} dx}\)

Answer 12

First, evaluate the integral and see wat u get,
u need to use substitution to evaluate the integral

Answer 13

dont wry about the limit yet...
first evaluate the integral

Answer 14

Oh, I used substitution & got.
u = -x^4
du = -4x^3

Answer 15

\[\frac{ -1 }{ 4 } \int\limits_{}^{} e ^{-u} du\]

Answer 16

u = -x^4
du = -4x^3 dx

-du/4 = x^3 dx

Answer 17

^^

Answer 18

yes ! that looks good... keep going...

Answer 19

Yes that, sorry I'm reading my paper too fast. lol

Answer 20

:) u need to setup the bounds also ok ?

Answer 21

Ok, so [0, infinity] its \[\frac{ -1 }{ 4 }e ^{-x^4}(-4x ^{3}) +C \] ?

Answer 22

\(\large \mathbb { \int\limits_{-\infty}^{0} x^3e^{-x^4} dx = \lim \limits_{a \rightarrow -\infty} \int\limits_{a}^{0}x^3e^{-x^4} dx} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \lim \limits_{a \rightarrow -\infty} \int\limits_{a^4}^{0}e^{-u} (\frac{du}{4})} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \lim \limits_{a \rightarrow -\infty} \frac{1}{4} \int\limits_{a^4}^{0}e^{-u} du} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \lim \limits_{a \rightarrow -\infty} \frac{1}{4} -e^{-u}\Big|_{a^4}^{0}} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \lim \limits_{a \rightarrow -\infty} \frac{1}{4} [-e^{-0} + e^{-a^4} ]} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \lim \limits_{a \rightarrow -\infty} \frac{1}{4} [-1 + e^{-a^4} ]} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \frac{1}{4} [-1 + e^{-\infty} ]} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \frac{1}{4} [-1 + 0]} \)

\(\large \mathbb { ~~~~~~~~~~~~~~~~~~~ = \frac{-1}{4}} \)

Answer 23

^^complete solution above..
see if that makes some sense...

Answer 24

similarly the integral other side equals 1/4
both adding gives u -1/4 + 1/4 = 0
as it should be for an odd function

Answer 25

Ok, thank you! I'm just confused as to why the bound is [a^4, 0], instead of [a, 0]

Answer 26

good question :)

we substituted u = x^4

so,
as x -> a, u -> a^4
as x -> 0, u -> 0

Answer 27

Ohhhh, okay. I get it now. Thank you very much!

Answer 28

np :) try to do the other integral also... and see if u get 1/4.. good luck !