Question

how would you integrate 3+x^2e^x^3 i know how to integrate e^x^3 but not when it is multiplied by x^2.

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

well I can't say for sure but you could try thinking like the chain rule of derivate

Answer 3

the derivative of x^3 is 3x^2 so you can use substitution u=x^3

Answer 4

rsvitale is right...

Answer 5

what do i do with the subsititution u?

Answer 6

do i have to use integration by parts of chain rule?
and why use the x^3 and not x^2?

Answer 7

sorry i ment OR chain rule

Answer 8

du=3x^2dx
1/3du=x^2dx

so you have:
\[\int\limits_{}^{}3+\frac{1}{3}\int\limits_{}^{}e^udu\]

Answer 9

now you have integrals you can compute

Answer 10

use u=e^x^3,,du=(e^x^3)3x^2

Answer 11

first integral wrt x

Answer 12

ahh im confused! i have to solve this:\[\int\limits 3 + e ^{x ^{3}}x ^{2}\] and my answr says \[3x+1/3e ^{x ^{3}}+C\]

can someone please explain how the x^2 completely dissapears? sorry im so confused and have been styuck on the same question for about 4 hours litterally! thanks :)

Answer 13

integral of (3+x^2e^x^3)dx =(e^x^3)/3 + 3x dx its perfect there with 1/3

Answer 14

use u=e^x^3,,du=(e^x^3)3x^2....its perfect there with 1/3

Answer 15

what is the name of that rule you are using?
i will look it up as maybe that is why i am not getting it.. i just dont see how or where the integral of 2x dissapears to

Answer 16

i mean intregral of x^2 sorry

Answer 17

its just a substitution rule using u=e^x^3,,du=(e^x^3)3x^2.dx.

Answer 18

the x^2 did not disappear there..in du=(e^x^3)3x^2 dx..

Answer 19

integral of (3+x^2e^x^3)dx
= integral of[(3+(3/3)x^2e^x^3)]dx
=(e^x^3)/3 + 3x +C answer

Answer 20

did u get it clk?

Answer 21

ahhhhh im soooo sorry i still dont get it much.. i got it earlier when my lecturer explained and then i didnt when tackling it myself, and still am a little confused as do exactly how the method works.. bassically i dont get how you went from
integral of (3+x^2e^x^3)dx
to
= integral of[(3+(3/3)x^2e^x^3)]dx <-- how is this the same as the previous line? how come we have 3/3x^2e^x^3?

Answer 22

using the substitution rule we use u=e^x^3...,and..,du=(e^x^3)3x^2 dx ..
,,did you see from the du that 3 was not in the original problem.?
so we have to balance it with 1/3 from the original problem..

Answer 23

yeah i just about get that we are balancing it out with 1/3... but i dont get what we do with our u and du/dx.. do we then make x^2 = v and then dv = 2x and then do something with that?
Im so sorry im being so useless.. feel free to give up with me lol.. im sure i will get it if i keep going... not going to give up..... :/

Answer 24

no not like that in this type of situation... use
use u=e^x^3...,and..,du=(e^x^3)3x^2 dx ..
so that
integ 3 + intg (1/3)du
=3x + (1/3)u +C
=3x + (1/3)e^x^3 +C answer

Answer 25

alternate solution is
let du=3x^2 dx then
(1/3)du=x^2 dx
so that
integ 3 +integ (1/3)e^u du
=3x + (1/3)e^u +C
=3x + (1/3)e^x^3 +C answer

Answer 26

okay i get that now.. so we have u=e^x^3 then du will be e^x^3 multiplied by the differential of x^3 which is 3x^2 so of coarse integrating that takes u to u again.. but where does the x^2 go? coz surely the whole substitution is done because of the x^2 but i dont get how or where it goes and how or why it disappears

Answer 27

on the first solution we got:
the x^2 did not disapear there it became part of du=(e^x^3)3x^2 dx ..

Answer 28

on the first solution we got: the x^2 did not disapear there it became part of du=(e^x^3)3x^2 dx ..
so that
integ 3 + intg (1/3)du
=3x + (1/3)u +C
=3x + (1/3)e^x^3 +C answer

Answer 29

did you see this...integ 3 + intg (1/3)du
the x^2 became part of du there

Answer 30

i almost get it.... is that because du includes x^2 , (3x^2 which we make 1/3 x 3x^2 to make it x^2) then we know that integrating will give us u again? am i closer to understanding?
btw if i dont reply.. means i have fallen asleep.. i have a lecture this afternoon.. i will get on it as soon as im up.
thanks so much for ur help

Answer 31

yes you got it there now..lol

Answer 32

integral of du=u+C

Answer 33

but dont forget 1/3 to balance it out ...lol

Answer 34

aaaah okay i think i get it now.. its rather hard though!!! plus its at the end of a differential equation.... that wont always be the case tho when integrating two functions surely? i swear i have done it completely differently before when integrating two functions multiplied together..
oh well thanks anyway

Answer 35

ok then...good luck clk ....

Answer 36

i know now that you got it...lol

Answer 37

lol i feel very stupid... i just gave u a medal or whatever that thing is for a good answer :)