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- Thread starter whatisphysics
- Start date

In summary, eumyang attempted to use substitution to solve for the integral but had an issue before. They then tried using u = x^8 and got the same result as before. They then tried using u = x^8 and got the same result as before.f

- #1

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- #2

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Try using u = x^8

- #3

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Try using u = x^8

The x^8 from e^(x^8) or x^7?

- #4

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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

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Okay, I've now ended up with

\\frac{1}{7}\\int\\frac{e^u}{x^7} du

\\frac{1}{7}\\int\\frac{e^u}{x^7} du

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

What is the derivative of u = x^8 with respect to x?

- #7

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Can you show me how you would do it?

- #8

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[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

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[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!

- #10

Homework Helper

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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 x

So,

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

- #11

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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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