So I have to prove the following:
If $f$ is monotone in an interval $I:\{a\leq x\leq b\}$, $\int_a^bf(x)dx$ is between $f(a)(b-a)$ and $f(b)(b-a)$.
I clearly see the geometric interpretation of this. The previous statement is saying that the area represented by the integral is "between" the area of a rectangle with height $f(a)$ and base $(b-a)$, and a rectangle with height $f(b)$ and base $(b-a)$. Obviously the function could be strictly increasing, decreasing or just being a constant (as far as I know that's the definition of a monotone function), as such I can visualize that the statement holds, however I don't seem to be able to put this into math notation and actually prove it.
I have tried to use a Riemann sum to solve it, I've also tried to use upper and lower integrals but I didn't really know what to do with those. Any help or insight on how to tackle this would be truly appreciated.
EDIT 1:
So, from the comment I came up with the following (for the case in which we have a strictly increasing function):
If $f(b)>f(a)$, then $\int_a^b[f(b)-f(a)dx]\geq 0$, since the integrand is positive, while $\int_a^b[f(a)-f(x)]dx\leq 0$, because the integrand is negative. From this we get the following inequality:
$\int_a^b[f(a)-f(x)]dx\leq 0\leq\int_a^b[f(b)-f(a)]dx$
By using properties of the integrals, we simplify as: $\int_a^bf(a)-\int_a^bf(x)dx\leq 0\leq\int_a^bf(b)dx-\int_a^bf(x)$
It is then obvious,
$\int_a^bf(a)\leq\int_a^bf(x)\leq\int_a^bf(b)$
For this last part I am not really sure how to convert the integrals of the function evaluated at the lower and upper bound of the integral into $f(a)(b-a)$ and $f(b)(b-a)$. I can see that the height will always be the same and the base is just going to be $(b-a)$. Maybe a Riemann sum of the previous integral would help? I have tried doing this:
$\int_a^bf(a)=\lim_{x\to \infty}\sum_{k=1}^nf(\xi)\Delta_kx$. Since we have that for every $\xi$, $f(\xi)$ will be the same as $f(a)$, then by properties of limits and sums (because f(a) is a constant), $\lim_{n\to \infty}\sum_{k=1}^nf(a)\Delta_kx=f(a)\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$, and since the sum of all the little sub-intervals that compose $\lim_{n\to \infty}\sum_{k=1}^n\Delta_kx$ is $(b-a)$, we finally have $f(a)(b-a)$. We then do the same for $f(b)$. This leads us to,
$f(a)(b-a)\leq\int_a^bf(x)\leq f(b)(b-a)$
- 1$\begingroup$ $ \int_a^b f(x) \, dx \le \int_a^b f(b) \, dx =$ ...? $\endgroup$Martin R– Martin R2021-03-16 05:50:38 +00:00Commented Mar 16, 2021 at 5:50
3 Answers
If $f(b)> f(a),$ then $\int_a^b (f(b)-f(x)) d x \geq 0,$ since the integrand is positive, while $\int_a^b (f(s)-f(x)) d x \leq 0,$ since the integrand is negative.
If $f$ is increasing, then by the mean value theorem for integrals, there is $ \mu \in [f(a),f(b)]$ such that
$$\int_a^b f(x)dx= \mu(b-a).$$
Since $f(a) \le \mu \le f(b)$, we have $f(a) (b-a) \le \mu(b-a) \le f(b)(b-a)$ and the result follows.
If $f$ is decreasing, then by the mean value theorem for integrals, there is $ \mu \in [f(b),f(a)]$ such that
$$\int_a^b f(x)dx= \mu(b-a).$$
Since $f(b) \le \mu \le f(a)$, we have $f(b) (b-a) \le \mu(b-a) \le f(a)(b-a)$ and the result follows.
- $\begingroup$ The proof of the mean value theorem for integrals (at least the one that I know of) uses the fact that $\int_a^b f(x) dx$ lies between $m(a-b)$ and $M(a-b)$, where $m$ and $M$ are the minimum and maximum of $f$ on the interval. In that case, your argument would be a circular reasoning. $\endgroup$Martin R– Martin R2021-03-16 08:07:34 +00:00Commented Mar 16, 2021 at 8:07
Here is a proof using Riemann sums.
Since $f$ is Riemann integrable, for any $\epsilon > 0$ there is a partition $P_\epsilon: a = x_0 < x_1 < \ldots < x_n =b$ such that for any choice of tags $\xi_j \in [x_{j-1}, x_j]$, the Riemann sum $S(P_\epsilon,f)$ satisfies
$$\tag{1}\int _a^bf(x) \, dx - \epsilon \leqslant S(P_\epsilon,f) = \sum_{j=1}^n f(\xi_j) (x_j-x_{j-1}) \leqslant \int_a^bf(x) + \epsilon,$$
Assume WLOG that $f$ is monotone increasing (otherwise we can consider $-f$). We have $f(a) \leqslant f(x) \leqslant f(b)$ for all $x \in [a,b]$, and it follows that for any Riemann sum
$$\tag{2}f(a)(b-a) = \sum_{j=1}^n f(a)(x_j-x_{j-1})\leqslant \underbrace{S(P,f)}_{\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1})} \leqslant \sum_{j=1}^n f(b)(x_j-x_{j-1})\leqslant f(b)(b-a) $$
Supoose $\int_a^b f(x) \,dx > f(b)(b-a)$. Taking $0 < \epsilon < \int_a^b f(x) \, dx - f(b)(b-a)$, it follows from (1) that
$$f(b)(b-a) < \int_a^b f(x) \, dx - \epsilon \leqslant S(P_\epsilon,f)$$
However, as (2) holds for any partition and corresponding Riemann sum we get the contradiction
$$S(P_\epsilon,f) \leqslant f(b)(b-a) < S(P_\epsilon,f)$$
By a similar argument we get a contradiction if $\int_a^b f(x) \,dx < f(a)(b-a)$ and, therefore, we conclude
$$f(a)(b-a) \leqslant \int_a^b f(x) \, dx \leqslant f(b)(b-a)$$
