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anonymous

  • 4 years ago

If F(x)= the integral of f(u)du from 1 to x^3-10 and f(-2)=5, the F'(2)=?

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  1. anonymous
    • 4 years ago
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    \[F(x)=\int\limits_{1}^{x^3-10}f(u)du\]

  2. anonymous
    • 4 years ago
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    \[F(x)=\int\limits_{1}^{x^3-10} f(u) du\] and f(-2)=5, the F'(2)=?

  3. JamesJ
    • 4 years ago
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    Hence if one of the antiderivative of f(x) is \[ G(x) = \int f(x) \ dx \] then by the Fundamental Theorem of Calculus, \[ \frac{dG}{dx} = f(x) \] Now, given the definition of F(x), we have that \[ F(x) = G(x^3-10) - G(1) \] therefore

  4. JamesJ
    • 4 years ago
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    ...therefore \[ F'(2) = \frac{d \ }{dx} ( G(x^3 - 10) - G(1) )|_{x=2} = \frac{d \ }{dx} G(x^3-10) |_{x=2} \]

  5. JamesJ
    • 4 years ago
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    So you need to find that derivative and evaluate it at x = 2.

  6. JamesJ
    • 4 years ago
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    Hint: use the chain rule and the Fund Thm of Calculus result above.

  7. anonymous
    • 4 years ago
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    I'm a bit confused. I know F'(x) = f(x) Therefore making F'(2) = f(2), and f(u) would be the integrand, but where would I use the given f(-2)=5

  8. anonymous
    • 4 years ago
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    maybe to find out to C

  9. anonymous
    • 4 years ago
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    It's a definite integral though, so I don't think it has a C. :/

  10. JamesJ
    • 4 years ago
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    F'(x) = (d/dx)G(x^3-10) = 3x^2 . G'(x^3-10) , by the chain rule. =3x^2.f(x^3-10) , by the Fundamental Theorem of Calculus Hence F'(2) = 12.f(-2) Now ... I'm sure you see it.

  11. anonymous
    • 4 years ago
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    ...wait I just checked the answer key and it says the answer is 60, but I don't understand how

  12. JamesJ
    • 4 years ago
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    Yes, exactly. You're given that f(-2) = 5, hence \[ F'(2) = 12.f(-2) = 12(5) = 60 \]

  13. JamesJ
    • 4 years ago
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    So again. Suppose that an anti-derivative (not "the antiderivative", there's no such thing because there's a constant of integration) of f(x) is G(x). Then by the Fundamental Theorem of Calculus, \[ G'(x) = f(x) \] Now the function F is given by \[ F(x) = \int^{x^3-10}_1 f(u) \ du = G(x^3 - 10) - G(1) \] Therefore \[ F'(x) = \frac{d \ }{dx}G(x^3-10) - \frac{d \ }{dx} G(1) \] That second term is zero because \( G(1) \) is a constant. Using the chain rule on the first term, \[ \frac{d \ }{dx}G(x^3-10) = \frac{d \ }{dx}(x^3-10)G'(x^3-10) \] \[ = 3x^2.G'(x^3-10) \] \[ = 3x^3.f(x^3-10) \ \ \text{, by the Fundamental Theorem of Calculus} \]

  14. JamesJ
    • 4 years ago
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    Therefore \[ F'(2) = 3(2)^2 . f(2^3-10) \] \[ = 12.f(-2) \] \[ = 12(5) \ \ \text{, because we are given that } f(-2) =5 \] \[ = 60. \]

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