# Approximating area under/above the curve

I got this problem and not sure how to relate the derivative and the integral:

Let $f : [0, 1] \rightarrow \Bbb R$ be twice differentiable, and assume that $f(0) = f(1) = 0$, $f''$ is continuous, and $|f''(x)| \leq 1$ for all $x \in [0, 1]$. Show that $$\left | \int_0^1 f(x) \space dx \right | \leq {1 \over 12}.$$

(Hint: Consider $\int_0^1 \left ( {x^2 \over 2} - {x \over 2} \right )f''(x) \space dx$.)

By analyzing what is giving, $f(0) = f(1) = 0$ in particular, we can see that $f$ got at least one maximum/minimum somewhere on $(0, 1)$, or even none if it's completely $f(x) = 0$.

The way I interpret the problem, is that if both function's ends are tied at $0$, show that whatever the curve of the function line might be and as long as the function has a continuous second derivative with absolute value at most 1, the area under/above the curve is at most $1 \over 12$.

Sounds pretty clear, but how to approach the problem is the question.

I don't understand what the hint tries to say.

Could someone explain me a solution or at least give a more understandable hint on how to solve this?

• Do you believe that $0<f(x)<g(x)$ implies $\int 0 \mathrm{d}x < \int f(x) \mathrm{d}x < \int g(x) \mathrm{d}x$ for $f$ and $g$ satisfying the hypotheses of the problem? Once you do, try integrating $-1 < f'' < 1$ twice, applying the boundary conditions to fix the constants of integration (separately for each side of the inequality) and then notice what you got. – Eric Towers Apr 27 '14 at 2:38
• Note that the integral of $x^2/2-x/2=-1/12$, which is probably relevant. Also, if you perform integration by parts on the integral in the hint twice, then you can relate this to the integral of $f$. – Mark McClure Apr 27 '14 at 2:43

$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx = \dots = \int_{0}^{1}f(x)\, dx$$
$$\left|\int_{a}^{b}f(x)g(x)\,dx\right| \leq \int_{a}^{b}|f(x)||g(x)|\, dx$$
• I integrated the hint by parts and got the same equality, and found the theorem about absolute value of an integral. I also see that $\int_0^1 \left | {x^2 \over 2} - {x \over 2} \right | = {1\over 12}$ and that $\left |f''(x) \right | \leq 1$, but how do I get $\int_0^1 \left | {x^2 \over 2} - {x \over 2} \right | \left |f''(x) \right | \leq \left | -{1 \over 12} \right | |1|$? – Katestrophical Apr 27 '14 at 4:29
• $\int_{0}^{1}\left|\frac{x^{2}}{2}-\frac{x}{2}\right||f^{\prime\prime}(x)| \, dx \leq \int_{0}^{1}\left|\frac{x^{2}}{2}-\frac{x}{2}\right|\times 1 \, dx = 1/12$. – JessicaK Apr 27 '14 at 5:28