## 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)=?

1. anonymous

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

2. anonymous

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

3. JamesJ

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

...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

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

6. JamesJ

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

7. anonymous

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

maybe to find out to C

9. anonymous

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

10. JamesJ

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

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

12. JamesJ

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

13. JamesJ

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

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