Find the maximum difference between the limits of integration

Question

Let $ C=\int_{a}^{b} (7x-x^2-10) dx $ where $a<b$

Determine the maximum value $(b-a)$ can assume if $C=0$

This question has troubled me for some time, and I would like some help to solve this problem. The correct answer is $(b-a)≈5.2$

I plotted the graph of the quadratic polynomial and found that $a=0.5$ and $b=5$ leads to $C=0$ Here is the picture of the graph

But this means that $(b-a)=4.5$ which is the wrong answer

Answer 1

We have that

\begin{align*} \int_a^b(7x-x^2-10)\,\mathrm{d}x &=\biggl[\frac{7}{2}x^2-\frac{1}{3}x^3-10x\biggr]_a^b \\ &=\frac{7}{2}(b^2-a^2)-\frac{1}{3}(b^3-a^3)-10(b-a) \\ &=(b-a)\left(\frac{7}{2}(b+a)-\frac{1}{3}(a^2+ab+b^2)-10\right) \\ &=0. \end{align*}

Make now the change of variables

$$\begin{cases} u=b-a,\\ v=b+a, \end{cases}$$

so that

\begin{align*} 0 &=\frac{7}{2}(b+a)-\frac{1}{3}(a^2+ab+b^2)-10 \\ &=\frac{7}{2}v-\frac{1}{3}\left(v^2-\frac{1}{4}(v-u)(v+u)\right)-10 \\ &=\frac{7}{2}v-\frac{1}{3}\left(v^2-\frac{1}{4}(v^2-u^2)\right)-10 \\ &=-\frac{1}{12}u^2-\frac{1}{4}v^2+\frac{7}{2}v-10. \end{align*}

Rearranging we get that

$$u=\sqrt{-3v^2+42v-120}=\sqrt{27-3(v-7)^2}$$

(note that $u=b-a>0$ as we can assume $b>a$, so we take the positive root). From this it is clear that the biggest value of $u=b-a$ is $\sqrt{27}=3\sqrt{3}$.

Answer 2

Another way to do it

Let $b=c+a$ which gives for the integral $$I=-\frac c 6\left(2c^2+3 (2 a-7)c+6 (a-5) (a-2)\right)$$ Just a quadratic equation in $c$ whose roots are $$c_\pm=\frac{1}{4} \left((21-6a)\pm \sqrt{3(-4 a^2+28 a-13)}\right)$$ Considering $c_+$ and differentiating $$(c_+)'=\frac{1}{4} \left(-6+\frac{\sqrt{3} (28-8 a)}{2 \sqrt{-4 a^2+28 a-13}}\right)$$ Again a quadratic equation $$(c_+)'=0 \quad \implies \quad a=\frac{7-3 \sqrt{3}}{2} $$ and, for this value of $a$, $$c_+=3 \sqrt{3}$$