# Thread: Integrals and Proofs

1. ## Integrals and Proofs

Hey there,

Here's the question:

a.) Prove that if f is integrable on [a,b] and m <= f(x) <= M for all [a,b], then:
$
\int_a^b f(x) dx = (b-a)k
$

for some number "k" with m <= k <= M

b.) Prove that if f is continuous on [a,b], then
$
\int_a^b f(x) dx = (b-a)f(q)
$

for some q in [a,b]

So for the work I set up that U(f,P) - L(f,P) < E, which implies the usualy equasion of the Sum of (Mi-mi)(Change in ti) < E. But I'm unsure where I should start throwing in this "k" variable? Thanks

2. Consider the function

$F(x)=\int_{a}^{x}f(t)dt$

First note that $F(x)$ is differentiable (why)

Now use the mean value theorem on $[a,b]$

$F(b)-F(a)=F'(c)(b-a)$

Can you finish from here

3. F(x) is differentiable because we have assumed that the function f(t) is integrable?
From there, using the mean value theorem I would see that in the end we'd get

$
\int_{a}^{b}f(t)dt - \int_{a}^{a}f(t)dt = F'(c)(b-a)$

$
\int_{a}^{b}f(t)dt = F'(c)(b-a)
$

Does this end up meaning F'(c) = k? and thus we have our result?

4. Remember what the function

$F(x)=\int_{a}^{x}f(t)dt$ represents and you still need to show that
$m < k < M$

What is $F'(x)$ equal to?

So the question is why is $m< F'(c) < M$ for some $c \in (a,b)$

5. Would F'(x) = f(x), then? So F'c = f(c). In this case does this f(c) become our k? I'm still having some trouble trying to figure out why exactly F'(c) does stay between m and M. Could I say f(c) (or k) would have to be between m and M due to previous theorems regarding continuity (if f(a) < c < f(b), there exists some x with f(x) = c? and that we do obviously have some bounds?

6. Originally Posted by mscbuck
Would F'(x) = f(x), then? So F'c = f(c). In this case does this f(c) become our k? I'm still having some trouble trying to figure out why exactly F'(c) does stay between m and M. Could I say f(c) (or k) would have to be between m and M due to previous theorems regarding continuity (if f(a) < c < f(b), there exists some x with f(x) = c? and that we do obviously have some bounds?
hmm. I think that this proof will only work for part b when f is continuous.

The function $F(x)=\int_{a}^{x}f(t)dt$ can only be differentiated by the FTC when $f(t)$ is continuous.

So here is a different proof:
Since
$\displaystyle m < f(x) < M \implies \int_{a}^{b}mdx < \int_{a}^{b}f(x)dx< \int_{a}^{b}Mdx$

$\displaystyle m(b-a) < \int_{a}^{b}f(x)dx < M(b-a) \implies m < \frac{\int_{a}^{b}f(x)dx}{b-a} < M$

Now choose $\displaystyle k=\frac{\int_{a}^{b}f(x)dx}{b-a}$

7. Thanks EmptySet! I actually figured out part a. right before you posted, but I did the exact same steps so I'm glad. An add-on to this is a question that asks "Show by an example that continuity is essential". I'm unsure of what he is asking, because I know that we can have discontinuous functions that still can be integrated. Continuity is sufficient but not necessary for integrability right?

8. Originally Posted by mscbuck
Thanks EmptySet! I actually figured out part a. right before you posted, but I did the exact same steps so I'm glad. An add-on to this is a question that asks "Show by an example that continuity is essential". I'm unsure of what he is asking, because I know that we can have discontinuous functions that still can be integrated. Continuity is sufficient but not necessary for integrability right?
Yes that is correct consider the function

$\displaystyle f(x)=\begin{cases} 0 \text{ if } 0 \le x \le \frac{1}{2} \\ 1 \text{ if } \frac{1}{2} < x \le 1 \end{cases}$

and the integral $\displaystyle \int_{0}^{1}f(x)dx$