Question

Can someone please help me with the beginning of this integral:

Answer 1

\[\int\limits_{0}^{1} x^3e^{x^3}dx\]

Answer 2

\[\int_{0}^{1} x^3e^{x^3}dx\]
\[\int_{0}^{1} xx^2e^{x^3}dx\]
\[\frac{1}{3}\int_{0}^{1} x~(3x^2e^{x^3})dx\]
it looks to be a rather simple by parts now

Answer 3

u = x^3

u^(1/3) = x

du/3u^(2/3) = dx

\[\int ue^udx\]
\[\frac{1}{3}\int u^{-1/3}e^udu\]if you wanna get fancy

Answer 4

the problem with that substitution is that wouldn't you be stuck with an integral that is even harder to deal with

Answer 5

perhaps, and looking at the wolf, it includes a gamma function

Answer 6

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

Answer 7

yeah, gamma function: http://www.wolframalpha.com/input/?i=integrate+%28x%5E3%29*%28e%5E%28x%5E3%29%29+from+0+to+1

Answer 8

lol

Answer 9

is it spose to be this level of difficulty? or are you working at a high level to begin with?

Answer 10

1st year university maths course

Answer 11

is the problem written correctly? or is it possible: x^2 e^(x^3) ??

Answer 12

Maybe there's a simpler way we're not thinking of, or the problem was copied wrong. Are you sure it's not\[\int\limits_{0}^{1}x^2e^{x^3}dx\]
because that would be much better lol

Answer 13

nope, it's definitely: (x^3) * (e^(x^3))

Answer 14

remnant have you really learned any other techniques than u-substitution? I'm just wondering what kind of toolset you have to play with lol.

Answer 15

maybe it was meant as a hard problem to work through. thanks for the help though. just gonna leave it for now as it does seem a lot harder than what I'm accustomed to.

Answer 16

Good luck have fun.

Answer 17

\[e^u=1+\frac{u}{1!}+\frac{u^2}{2!}+\frac{u^3}{3!}+~...\]
\[e^{x^3}=1 +\frac{x^3}{1!}+\frac{x^6}{2!}+\frac{x^9}{3!}+~...\]
\[x~e^{x^3}=x +\frac{x^4}{1!}+\frac{x^7}{2!}+\frac{x^{10}}{3!}+~...\]
integrating that gives me
\[\int x~e^{x^3}=\frac{1}{2}x^2 +\frac{1}{5}\frac{x^5}{1!}+\frac{1}{8}\frac{x^8}{2!}+\frac{x^{10}}{3!}+~...\]

is one thought i have, thats prolly wrong tho ;)

Answer 18

at x=0, thats 0, so at x=1

\[\frac{1}{2}+\frac{1}{5}+\frac{1}{16}+\frac{1}{240}+...\]

Answer 19

11, not 10 :/

Answer 20

\[\sum_{n=1}^{inf}\frac{1}{(3n-1)n!}\]

Answer 21

\[\int\limits_{0}^{1}x^3e^{x^3}dx\]
\[\frac{1}{3}\int\limits_{0}^{1}(x)3x^2e^{x^3}dx\]
u = x^3
du = 3x^2dx
du/3x^2 = dx
insert

\[\frac{1}{3}\int\limits_{0}^{1}(x)e^{u}du\]

then

\[\frac{1}{3} \int\limits_{0}^{1}u^{\frac{1}{3}}e^udu\]

blah this would be hard to deal with, maybe you can represent it as a series and solve it that way?

Answer 22

Thanks for all the help :)

Answer 23

why did we integrate xe^(x^3) ?\[u=x\implies du=dx\]\[dv=3x^2e^{x^3}dx\implies v=e^{x^3}\]\[\int_0^1x^3e^{x^3}dx=\left.\frac13xe^{x^3}\right|_0^1-\frac13\int_0^1e^{x^3}dx=\frac e3-\frac13\int_0^1e^{x^3}dx\]

Answer 24

\[x~e^{x^3}=x +\frac{x^4}{1!}+\frac{x^7}{2!}+\frac{x^{10}}{3!}+~...\]
forgot it was x^3 and not x :)

\[x^3~e^{x^3}=x^3 +\frac{x^6}{1!}+\frac{x^9}{2!}+\frac{x^{12}}{3!}+\frac{x^{15}}{4!}+~...\]

\[\int x^3~e^{x^3}=\frac{x^4}{4} +\frac{x^7}{7}+\frac{x^{10}}{20}+\frac{x^{13}}{13*3!}+\frac{x^{16}}{16*4!}+~...\]

at x=0 same concept, zeros out

at x=1 we have\[\sum_{n=1}^{inf}\frac{1}{n!(1+3n)}\]

might prove better

Answer 25

my summation was not matching the integral, and i figured out why.

i need to start with n! = 0! and at the moment im starting off at n!= 1! which is throwing my results off.\[\sum_{n=1}^{inf}\frac{1}{(n-1)!(1+3n)}\]that would work out fine, but it looks to crowded for me so lets adjust it.\[\sum_{n=1-1}^{inf}\frac{1}{(n-1+1)!(1+3(n+1))}\]
\[\sum_{n=0}^{inf}\frac{1}{n!(4+3n)}\]And the wolf finally agrees that that is a match :)

http://www.wolframalpha.com/input/?i=sum+%28n+%3D+0+to+inf%29+1%2F%28n%21%283n%2B4%29%29