# Prove that $\left|\int_a^b f(x) dx - (b-a) f(a) \right| \leq \frac{(b-a)^2}{2}$ when $f$ satisfies $|f(u)-f(v)|\leq |u-v|$

this is a question from Apostol's Calculus, Volume I, p. 139 (Exercises 33) which I've gotten completely stuck on.

Let $f$ be a continuous function such that $|f(u) - f(v)| \leq |u - v|$ for all $u$ and $v$ in an interval $[a,b]$.

(b) Assume that $f$ is integrable on $[a,b]$. Prove that $$\left| \int_a^b f(x) dx - (b-a) f(a)\right| \leq \frac{(b-a)^2}{2}.$$

In part (a) we proved that $f$ was continuous at each point of $[a,b]$. This proof is in the section on the basic limit theorems and definition of continuity, so essentially the only knowledge we have is of the definitions and basic theorems regarding limits. We have also proved the continuity of indefinite integrals (though it is isn't clear to me how or if to apply that theorem).

Geometrically, I believe I understand in a non-rigorous way what is happening, and why it is true. The quantity on the left is the area between $f(x)$ and the horizontal line at $f(a)$ from $a$ to $b$. The quantity on the right is the area of the right triangle of base and height $b-a$ (or viewed as an integral it is $\int_a^b x dx$). The condition on $f$ we are given in the hypothesis of the theorem roughly states that the change in $f$ over any interval must be less than the change in $x$ (or the width of the interval). So, we have a function $f$ that is changing less than linearly over $[a,b]$; hence, the integral of the function will be $\leq$ than that of $f(x) = x$.

I cannot figure out at all how to turn this intuition/geometric argument into a rigorous analytical one. Any help is appreciated. Thanks.

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Hint: Notice that $$\int_{a}^{b}f(x)dx-\left(b-a\right)f(a)=\int_{a}^{b}\left(f(x)-f(a)\right)dx.$$

Using the fact that $|f(v)-f(u)|\leq |v-u|$, the above is bounded in absolute value by

$$\leq\int_{a}^{b}|x-a|dx=\int_{a}^{b}(x-a) dx$$ since $x\geq a$ on this interval. Can you solve it from here?

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Yes! Fantastic. For some reason banging my head against a wall for a while did not cause me to notice that I could bring that $f(a)$ inside the integral... – user23784 Mar 4 '12 at 2:30

### Another way to solve this problem:

Note that $$\lim\left| \dfrac{f(x)-f(y)}{x-y}\right|\le 1,\,\forall x,y\\ \implies \lim_{x \to y}\left| \dfrac{f(x)-f(y)}{x-y}\right|\le 1,\,\forall y \\ \implies |f'(y)|\le 1, \forall y$$ Now define $F(x)=\int_{a}^{x}f(t)\,dt$.

Now note that \begin{align} \int_{a}^{b}f(t)\,dt&=F(b) \\&=F(a)+(b-a)F'(a)+\frac{(b-a)^2}{2!}F''(c)\\&=0+(b-a)f(a)+\frac{(b-a)^2}{2!}f'(c) \end{align}

So, $$\left|\int_a^b f(t) dt - (b-a) f(a)\right|\\=\frac{(b-a)^2}{2}|f'(c)|\le \frac{(b-a)^2}{2}$$

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