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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)

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

$$\implies -2x <xf'(x)< 2x$$ 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.

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

enter image description here

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  • $\begingroup$ 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? $\endgroup$ – Abcd Dec 17 '18 at 5:01
  • $\begingroup$ And I am not sure i completely understand the graphical solution. $\endgroup$ – Abcd Dec 17 '18 at 5:04
  • $\begingroup$ @Abcd: I did definite integration. $f(0) = 0$ is given, therefore $f(x) = \int_0^x f'(t)dt < 2x$. $\endgroup$ – Martin R Dec 17 '18 at 5:43
  • $\begingroup$ @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? $\endgroup$ – Martin R Dec 17 '18 at 6:05

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