Evaluating point using derivative graph

[wvcaudill2]

Homework Statement

*note that this is the graph of the derivative of the function f(x).

Homework Equations

Possibly the fundamental theorems of calculus?

The Attempt at a Solution

I know this problem should be easy, but I am not seeing how my book comes to the answer. Below is the solution my book gives:
f(-2)=f(2) + $$\int$$f'(x)dx = 1-$$\frac{1}{2}$$
The limits of integration are from 2 to -2

Obviously, the first part of the eqn is given, but what exactly does the integral do, and why should the limits be reversed? Furthermore, it seems that if you evaluated the eqn, you would end up with:
f(2) + f(-2) - f(2)
it seems like this would get you nowhere. . .

 

[Dick]

Well, first of all what is the integral of f'(x) from -2 to 2? Just add up the signed area under the curve of f'(x). Then use that the result should be f(2)-f(-2) by the fundamental theorem of calculus.

 

[LCKurtz]

You know the fundamental theorem of calculus, if F is an antiderivative of f then

$$F(b) - F(a) = \int_a^b f(x)\, dx$$

The point being that the integrand, f, is the derivative of the function F. Rephrasing it in those terms gives the equivalent form

$$f(b) - f(a) = \int_a^b f'(x)\, dx$$

Using -2 and 2 for a and b:

$$f(2) - f(-2) = \int_{-2}^2 f'(x)\, dx$$

and solving for f(-2) gives your equation

$$f(-2) = f(2) - \int_{-2}^2 f'(x)\, dx$$

Now to solve the problem you have only to realize that integrals geometrically represent signed area under the curve. So look at the picture and total up that area.

 

[wvcaudill2]

Oh, I was unaware of the equivalent form. This makes much more sense now!

One last question though. One of the theorems states that the derivative of an integral is the original function. Does this statement also work in reverse? I.e. the integral of a derivative also yields the original equation?

 

[Dick]

Oh, I was unaware of the equivalent form. This makes much more sense now!

One last question though. One of the theorems states that the derivative of an integral is the original function. Does this statement also work in reverse? I.e. the integral of a derivative also yields the original equation?

Yes, the integral of a derivative is the difference of the original function.

 

[LCKurtz]

Oh, I was unaware of the equivalent form. This makes much more sense now!

One last question though. One of the theorems states that the derivative of an integral is the original function. Does this statement also work in reverse? I.e. the integral of a derivative also yields the original equation?

To within a constant. Start with f(x) = x² + 5. Its derivative is 2x. When you take the integral of 2x you get x² + C because you have no way of knowing the value of any constant that may have been added to x² before you differentiated it.

 

[wvcaudill2]

Okay, thanks for your help guys! I really appreciate it!