# Maximum and minimum value of $\int_0^1 f(x)dx$ given $|f'(x)|<2$

Let $$f:\mathbb R\to \mathbb R$$ be a differentiable function such that $$f(0)= 0$$ and $$f(1)= 1$$ and $$|f'(x)|<2 ~ \forall x \in \mathbb R$$, if $$a$$ and $$b$$ are real numbers such that the set of possible values of $$\displaystyle\int_0^1 f(x)dx$$ is the open interval $$(a,b)$$, then $$b-a$$ is: ?

Attempt:

$$I = \int_0^1 1.f(x) dx$$

$$\implies I = 1 - \int_0^1 xf'(x)dx$$ (Integration by parts)

$$-2 < f'(x) < 2$$

$$\implies -2x for $$x>0$$

$$\implies -1 < \int_0^1 xf'(x) dx < 1$$

Therefore $$I_{max} = 2$$ and $$I_{min} = 0$$

$$\implies b- a = 2$$ but answer given is $$b-a = \dfrac 3 4$$.

Please let me know my mistake, and the correct way to solve it.

• I'm guessing the problem statement says $f(x)$ is nonnegative? – user25959 Nov 2 '18 at 19:26
• @user25959 it does not. – Abcd Nov 2 '18 at 19:32

Your estimates are correct, but not sharp. For example, in $$I = 1 - \int_0^1 xf'(x)dx > 1 - \int_0^1 2 x dx = 0$$ $$I$$ would be close to the lower bound $$0$$ only if $$f'(x) \approx 2$$ on the entire interval, which is not possible with $$f(0)=0$$ and $$f(1) =1$$.

The solution is actually simpler: Use the mean-value theorem to obtain upper and lower bounds for the admissible functions $$f$$.

• From $$f(0) = 0$$ and $$f'(x) < 2$$ it follows that $$f(x) < 2x$$ for $$0 < x \le 1$$.

• From $$f(1) = 1$$ and $$f'(x) > -2$$ it follows that $$f(x) < 1 - 2(x-1) = 3-2x$$ for $$0 \le x < 1$$.

Together: $$f(x) < \max(2x, 3-2x)$$ for $$0 < x < 1$$, which implies that $$\int_0^1 f(x) \, dx < \frac 78$$.

Similarly show that $$\int_0^1 f(x) \, dx > \frac 18$$.

Finally show that the integral can be arbitrarily close to those bounds.

You can also solve it graphically by drawing lines with slopes $$-2$$ and $$+2$$, starting from the given points $$(0, f(0))$$ and $$(1, f(1))$$.

The graph of $$f$$ must lie between the green and the red curve. $$(b-a)$$ is the area between those curves, and that is equal to the area of the blue rectangle.

• In $f'(x)< 2$ $\implies f(x)< 2x$, where is the constant of integration? Because you haven't preformed definite integration to both sides of the inequality right? – Abcd Dec 17 '18 at 5:01
• And I am not sure i completely understand the graphical solution. – Abcd Dec 17 '18 at 5:04
• @Abcd: I did definite integration. $f(0) = 0$ is given, therefore $f(x) = \int_0^x f'(t)dt < 2x$. – Martin R Dec 17 '18 at 5:43
• @Abcd: Which part of the graphical solution is unclear? That the graph of $f$ lies between the green and the red curve, that $(b-a)$ is the area between those curves, or that the area between the green and red curve is equal to the area of the blue rectangle? – Martin R Dec 17 '18 at 6:05