Stuck in Evaluating an Indefinite Integral

  • #1
whatisphysics
30
0

Homework Statement



Evaluate the indefinite integral.

Homework Equations


I tried to use substitution.


The Attempt at a Solution


I'm not so good with using the LaTex codes...so I attached a file I made in Word.

Thanks in advance!
 

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Answers and Replies

  • #2
iamalexalright
164
0
Try using u = x^8
 
  • #3
whatisphysics
30
0
Try using u = x^8

The x^8 from e^(x^8) or x^7?
 
  • #4
iamalexalright
164
0
I don't understand your question

But besides a poor choice of u in your original solution you don't carry out the integral properly

With my choice of u (u = x^8) compute du/dx and show me the new integral
 
  • #5
whatisphysics
30
0
Okay, I've now ended up with

\\frac{1}{7}\\int\\frac{e^u}{x^7} du
 
  • #6
iamalexalright
164
0
You should show me the steps you take to get there!

plus use the carrot to show superscripts with LaTeX: e^{u}

What is the derivative of u = x^8 with respect to x?
 
  • #7
whatisphysics
30
0
Can you show me how you would do it?
 
  • #8
iamalexalright
164
0
Alright, the integral is:

[tex]\int x^{7}e^{x^{8}}dx[/tex]

I'll choose u = x^8
[tex]du = 8x^{7}dx[/tex]

Here is where you had an issue before, notice that in the original integral I do have x^7dx in it ! This implies I don't need to divide by x^7, I just use the substitution:

[tex]\frac{1}{8}du = x^{7}dx[/tex]

Can you finish it from here?
 
  • #9
whatisphysics
30
0
Alright, the integral is:

[tex]\int x^{7}e^{x^{8}}dx[/tex]

I'll choose u = x^8
[tex]du = 8x^{7}dx[/tex]

Here is where you had an issue before, notice that in the original integral I do have x^7dx in it ! This implies I don't need to divide by x^7, I just use the substitution:

[tex]\frac{1}{8}du = x^{7}dx[/tex]

Can you finish it from here?

Okay, I tried what you said. Can you check it, thanks! And how do I go on from there, I'm not very good at this, but I still want to learn!
 

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  • #10
eumyang
Homework Helper
1,347
11
[tex]\begin{aligned}
dx &= \frac{du}{8x^7} \\
\int \frac{x^{7} \cdot e^u}{8x^{7}}dx &= ...
\end{aligned}[/tex]
This is the 2nd time I saw someone work a u-substitution this way. Why isolate dx? I think it makes things complicated. Solve for 'an expression in the integral that contains dx' instead, like this:
[tex]\begin{aligned}
u &= x^8 \\
du &= 8x^7 dx \\
\frac{1}{8} du &= x^7 dx
\end{aligned}[/tex]
(because x7 dx appears in the integral)

So,
[tex]\int x^{7}e^{x^{8}}dx = \frac{1}{8}\int e^u du = ...[/tex]
 
  • #11
whatisphysics
30
0
Thank you both so much!
And to eumyang, I don't understand, I thought it would be easier to isolate the dx, because you can simply just substitute the dx, isn't that right?

Anyway, thanks again!!
 

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