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anonymous
 4 years ago
If F(x)= the integral of f(u)du from 1 to x^310
and f(2)=5, the F'(2)=?
anonymous
 4 years ago
If F(x)= the integral of f(u)du from 1 to x^310 and f(2)=5, the F'(2)=?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[F(x)=\int\limits_{1}^{x^310}f(u)du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[F(x)=\int\limits_{1}^{x^310} f(u) du\] and f(2)=5, the F'(2)=?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Hence 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 \[ F(x) = G(x^310)  G(1) \] therefore

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1...therefore \[ F'(2) = \frac{d \ }{dx} ( G(x^3  10)  G(1) )_{x=2} = \frac{d \ }{dx} G(x^310) _{x=2} \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1So you need to find that derivative and evaluate it at x = 2.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Hint: use the chain rule and the Fund Thm of Calculus result above.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe to find out to C

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's a definite integral though, so I don't think it has a C. :/

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1F'(x) = (d/dx)G(x^310) = 3x^2 . G'(x^310) , by the chain rule. =3x^2.f(x^310) , by the Fundamental Theorem of Calculus Hence F'(2) = 12.f(2) Now ... I'm sure you see it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0...wait I just checked the answer key and it says the answer is 60, but I don't understand how

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, exactly. You're given that f(2) = 5, hence \[ F'(2) = 12.f(2) = 12(5) = 60 \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1So again. Suppose that an antiderivative (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 \[ F(x) = \int^{x^310}_1 f(u) \ du = G(x^3  10)  G(1) \] Therefore \[ F'(x) = \frac{d \ }{dx}G(x^310)  \frac{d \ }{dx} G(1) \] That second term is zero because \( G(1) \) is a constant. Using the chain rule on the first term, \[ \frac{d \ }{dx}G(x^310) = \frac{d \ }{dx}(x^310)G'(x^310) \] \[ = 3x^2.G'(x^310) \] \[ = 3x^3.f(x^310) \ \ \text{, by the Fundamental Theorem of Calculus} \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Therefore \[ F'(2) = 3(2)^2 . f(2^310) \] \[ = 12.f(2) \] \[ = 12(5) \ \ \text{, because we are given that } f(2) =5 \] \[ = 60. \]
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