Question

Another integral by substitution

Answer 1

\[\int\limits_{}^{}\sqrt[7]{e^x}dx\]

Answer 2

I don't think you can do this with substitution

Answer 3

Try integration by parts

Answer 4

stupid webworks, this is supposed to be the substitution section

Answer 5

are you writing the integral properly?

Answer 6

doesn't matter, don't get by parts either

Answer 7

7th root of e^x

Answer 8

@electrokid can you help me when you get a chance?

Answer 9

\[
u^7=e^x\implies7u^6du=e^xdx\implies{7u^6\over u}du=dx=7u^5
I=\int\sqrt[7]{u^7}(7u^5)du\\
I=7\int...
\]

Answer 10

@geerky42 math works in funny ways :)

Answer 11

\[\sqrt[7]{e^x} = e^{x/7}\]So let u = x/7. du = dx/7 \(\Rightarrow\) 7du = dx
\[7\int e^udu\]
I hope this helps.

Answer 12

that was good

Answer 13

Thanks.

Answer 14

Is this clear? @heradog

Answer 15

nope, totally lost
I understand making u=x/7 instead of e^x like I tried first
but after that it all went too fast for me

Answer 16

OK, we have u equals to x/7, right? We also need to substitute du because we cannot take integral of function with u in respect of x, so to find du, take derivative of u then take dx to the other side. Like this:\[u = \dfrac{1}{7}x \Rightarrow \dfrac{du}{dx} = \dfrac{1}{7} \Rightarrow du = \dfrac{1}{7}dx \Rightarrow7du = dx\]
Now we can substitute dx. Is this clear?

Answer 17

back into the original function?

Answer 18

no, I mean replace dx to 7du. Sorry.

Answer 19

then take 7 out of integral since it is constant.

Answer 20

\[
I=\int e^{x\over7}dx=\int e^u(7du)
\]

Answer 21

just put the words of geerky into math form -> the language herad understaands :)

Answer 22

Is it clear now? @heradog

Answer 23

almost, how do I get the du value putting x/7 back in?

Answer 24

you do not yet.
\[I=7\int e^udu=7e^u+C=7e^{x\over7}+C\\\boxed{I=7\sqrt[7]{e^x}+C}\]

Answer 25

You don't. du will disappear after taking integral. Don't worry about it. After integral, replace u back to x/7, just like what @electrokid did.

Answer 26

the integral cancels du and you just have the function part of it?

Answer 27

\[\int e^u du = e^u + C\] Plug x/7 into u because u = x/7. \[= e^{x/7}+C = \boxed{\sqrt[7]{e^x}+C}\]

Answer 28

Ok I think I get it

Answer 29

What part do you want me to clarify?

Answer 30

I understand the given problem once it's been all laid out, then I go to the next one which is slightly different and get confused again

Answer 31

Post another problem, we will help.

Answer 32

OK

Answer 33

\[\int\limits_{}^{}x^4e^{x^5}dx\]
so I want u to be x^5 because the derivative of that is (1/5)x^4

Answer 34

sorry derivative is 5x^4
antiderivative would be (1/6)x^6
I'm trying to combine them

Answer 35

Yes. so we can see that \(du = 5x^4dx\) but there is only \(x^4dx\) so take 5 to the other side, like this: \[\dfrac{1}{5}du = x^4 dx\]
Now substitute. \[ \dfrac{1}{5}\int e^u du\]

Can you handle it now?

Answer 36

\[\int x^4 e^{x^5}dx\]\[u = x^5 \space\space\space\space\space\space du = 5x^4 dx \Rightarrow \dfrac{1}{5}du = x^4 dx\]\[\int x^4 e^{x^5} dx = \dfrac{1}{5}\int e^udu\]

Answer 37

so how/when do you know to put e^u in for x^4

Answer 38

Well, if you take a close look at integral, you can see that the derivative of \(x^5\) is \(5x^4\) and integral does have \(x^4\), we replace it for du so we know we can use U-substitution. We do that to make integral easier to solve.

Answer 39

du=5x^4 dx
which means that
1/5du=x^4 dx
I follow that
how do you get from
1/5du=x^4 dx to
\[\frac{ 1 }{ 5 } \int\limits_{}^{}e^u du\]

Answer 40

\[\int x^4 e^{x^5}dx = \int\dfrac{1}{5}e^u du = \dfrac{1}{5}\int e^u du\]
Remember, \(\dfrac{1}{5}du = x^4 dx\)

Simply replace \(x^4 dx\) to \(\dfrac{1}{5}du\).

Is this clear? No?

Answer 41

OK I see it now!

Answer 42

You're welcome :)