Computing Coefficients of a Quadratic Equation given definite Integral

Question

For the quadratic function $$-ax^2 + 1$$ an upside down parabola with $y(0) = 1,$ is there a way to compute a such that the definite integral of $y$ between the roots ($x_1, x_2: f(x_1) \land f(x_2)= 0$) equals $1?$

Answer 1

the parabola is symmetric about $x = 0$ has roots $\pm \dfrac{1}{\sqrt a}$ so the area bounded by the parabola and $y = 0$ is $$\int_{-1/\sqrt a}^{1/\sqrt a}(1-ax^2) dx=2 \int_0^{1/\sqrt a} 1 - ax^2 dx = 2\left[ x - \dfrac{ax^3}{3}\right]_0^{1/\sqrt a}= 2(\dfrac{1}{\sqrt a} - \dfrac{1}{3\sqrt a}) = \dfrac{4}{3 \sqrt a}$$

if you set the area equal to one you can find $a.$

Answer 2

Hint: $y=-ax^2+1$ is zero when $x=\pm1/\sqrt{a}$, so it might be so that you want to calculate the integral $$ \int_{-1/\sqrt{a}}^{1/\sqrt{a}}1-ax^2\,dx. $$