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emz13

  • 2 years ago

How do i find the integral(or antiderivative) of (3-x)(x^2-6x)^5

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  1. myininaya
    • 2 years ago
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    let u=the inside of nastier part.

  2. myininaya
    • 2 years ago
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    Hello?

  3. emz13
    • 2 years ago
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    What do you mean?

  4. eashy
    • 2 years ago
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    u=x^2-6x du=2x-6=2(x-3)

  5. myininaya
    • 2 years ago
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    du=-2(3-x) dx -du/2=(3-x)dx

  6. emz13
    • 2 years ago
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    What do I do with the 3-x though?

  7. eashy
    • 2 years ago
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    replace (3-x)dx with -1/2 du

  8. emz13
    • 2 years ago
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    Why do I replace it with -1/2 du?

  9. eashy
    • 2 years ago
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    you want to convert the integral from integral(f(x)dx) to integral(f(u)du) if u=x^2-6x, then u'(x)=du/dx=2(x-3). thus du=dx u'(x)=2(x-3)dx thus -1/2*du=(3-x)dx the original integral is integrate( (3-x)(x^2-6x)^5 dx) =integrate( u^5 (3-x)dx) =integrate ( u^5*(-1/2) du) after integrating, convert back to x using the fact that u=x^2-6x

  10. emz13
    • 2 years ago
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    OH! okay I understand now. Thank you!!

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