- #1

- 479

- 4

^{4}) dx = [tex]\int_y^z[/tex] f(u) dx/du du, where u=x

^{4}

y=a

^{4}

z=b

^{4}

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- Thread starter squenshl
- Start date

- #1

- 479

- 4

y=a

z=b

- #2

HallsofIvy

Science Advisor

Homework Helper

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^{4}) dx = [tex]\int_y^z[/tex] f(u) dx/du du, where u=x^{4}

y=a^{4}

z=b^{4}

How you integrate the function f depends strongly on what f is, don't you think?

- #3

- 479

- 4

f is [tex]\int_-1^1[/tex] 1+x^{4} dx

- #4

- 479

- 4

meant to be a -1 at the bottom of the integral.

How could I do this then.

- #5

jgens

Gold Member

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[tex]\int_{-1}^1 1 + x^4\, dx[/tex]

Then it's pretty simple. Just seperate the integrand so that,

[tex]\int_{-1}^1 \,dx + \int_{-1}^1 x^4 \, dx[/tex]

But, based on the context of the question I'm not sure that's what you're looking for. Perhaps you could repost exactly what you mean clearly?

- #6

- 479

- 4

How do I evaluate this integral using the substitution u=x

[tex]\int_a^b[/tex] f(x

- #7

jgens

Gold Member

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- #8

- 479

- 4

I thought this was a more interesting way but just don't know how to get started.

- #9

- 479

- 4

How could I evaluate the integral above using u=x

- #10

Cyosis

Homework Helper

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Taking detours isn't a more interesting way, but here goes.

[itex]u=x^4 \Rightarrow du=4x^3dx[/itex]

[tex]

\int 1+x^4 dx=\int \frac{1+u}{4(u^{3/4})} du

[/tex]

Making life a lot harder!

[itex]u=x^4 \Rightarrow du=4x^3dx[/itex]

[tex]

\int 1+x^4 dx=\int \frac{1+u}{4(u^{3/4})} du

[/tex]

Making life a lot harder!

Last edited:

- #11

Gib Z

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- #12

Cyosis

Homework Helper

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Note: I didn't take the limits into account so you still have to do that yourself.

- #13

- 607

- 0

You are too harsh. Sure, for this function there are easy ways to do the integral. But for other functions f maybe you really should do the substitution. And why not learn to do the substitution on simple problems that can be evaluated in other ways? Then, for example, you can check your answers and see when you make a mistake.

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