Mean value theorem for integral

[jdz86]

Homework Statement

Suppose f is continuos on [a,b]. Show that there exists c in (a,b) such that the integral from a to b of f(x)dx equals (b-a)*f(c)

Homework Equations

The Attempt at a Solution

Tried using the mean value theorem to come up with a solution by rearranging different variables, but they don't relate cause it's f(c) not f '(c).

 

[gabbagabbahey]

You're not trying to show that $f(b)-f(a)=(b-a)f(c)$, you are trying to show $$\int_{a}^{b}f(x)dx=(b-a)f(c)$$.

Hint: suppose that the antiderivative of $f(x)$ is $F(x)$; what does the mean value theorem say about $F(b)-F(a)$? What does the fundamental theorem of calculus say about $F(b)-F(a)$?

 

[jdz86]

wow, ok definitely didn't look at it good enough. thanks a lot for the hint

 

[sutupidmath]

Or you could think of it this way:

SInce f is continuous on [a,b], then by the extreme value theorem it reaches its max and min value on that interval. That is there exist r,t such that f(r)=m, f(t)=M, where m and M are its smallest value and its greatest value on that interval.


THat is

$$m\leq f(x) \leq M$$ for all x in [a,b]

Now integrating from a to be we get

$$\int_{a}^bmdx\leq \int_{a}^bf(x)dx\leq \int_{a}^bMdx =>$$

$$m(b-a)\leq \int_{a}^bf(x)dx\leq (b-a)M$$

$$m\leq \frac{1}{b-a}\int_{a}^bf(x)dx\leq M$$

Now by the IVT there exists some number c on the interval (a,b) such that

$$f(c)=\frac{1}{b-a}\int_{a}^bf(x)dx=>\int_{a}^bf(x)dx=(b-a)f(c)$$

Proof done!