How does the discriminant of a quadratic equation give the definite integral?

Question

$$\int^7_1(x-1)(x-7)dx$$ Wee solve it by the usual method $$\large[\frac{x^3}3]_1^7+[7x]^7_1-[4x^2]^7_1$$ $$\frac{343}{3}-\frac13+49-7-196+4=-36$$ But if we take a look at the determinant, it is ${64-28}=36$. So taking the negative of it, we have the definite integral.
Is this just a coincidence?

Answer 1

It is definitely a coincidence. But there is an interesting historical remark to be made here. Upon shifting the lower root to $0$ and allowing the distance between the roots to be $a$ instead of $6$, we're looking at

$$\int_0^a x(a-x) dx.$$

You can get this with calculus if you want. However, the answer here was actually given by Archimedes long before we had modern calculus. He showed that the area between a downward parabola and the $x$ axis is $4/3$ times the area of the triangle formed by the two roots and the vertex. (He also geometrically showed the corollary when the chord is not horizontal.) In other words, in your case the area is $4/3$ times the area formed by the triangle with vertices $(0,0),(a,0),(a/2,a^2/4)$. This is $\frac{a^3}{8}$, so the area is $\frac{a^3}{6}$, which just happens to be equal to $a^2$ when $a=6$, which is what happens in your example.

Answer 2

$$\begin{align}\int_{a}^{b}(x-a)(x-b)dx&=\int_{a}^{b}(x-a)(x-a-(b-a))dx\\&=\int_a^b\left((x-a)^2-(b-a)(x-a)\right)dx\\&=\left[\frac{(x-a)^3}{3}-\frac{b-a}{2}(x-a)^2\right]_a^b\\&=-\frac{(b-a)^3}{6}\end{align}$$

The discriminant is $(-a-b)^2-4ab$.

Thus, we have$$-\frac{(b-a)^3}{6}=-(-a-b)^2+4ab$$$$\iff (b-a)^2(a-b+6)=0\iff b=a\quad\text{or}\quad b=a+6.$$

Answer 3

It appears to be a "coincidence" in the sense that $$\int_a^b(x-a)(x-b)dx\neq-(a-b)^2=-\Delta$$