Question

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

Answer 1

\[F(x)=\int\limits_{1}^{x^3-10}f(u)du\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

So you need to find that derivative and evaluate it at x = 2.

Answer 6

Hint: use the chain rule and the Fund Thm of Calculus result above.

Answer 7

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

Answer 8

maybe to find out to C

Answer 9

It's a definite integral though, so I don't think it has a C. :/

Answer 10

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.

Answer 11

...wait I just checked the answer key and it says the answer is 60, but I don't understand how

Answer 12

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

Answer 13

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,
\[ \frac{d \ }{dx}G(x^3-10) = \frac{d \ }{dx}(x^3-10)G'(x^3-10) \]
\[ = 3x^2.G'(x^3-10) \]
\[ = 3x^3.f(x^3-10) \ \ \text{, by the Fundamental Theorem of Calculus} \]

Answer 14

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