Question

The positive value of '$c$' that makes the area bounded by the graph of $y=c(1-x^2)$ and the x-axis equal to $1$, can be expressed in the form $p/q$ where $p,qϵN$ and in their lowest form, then $(p+q)$ equals:

Answer 1

$2{\int }_0^{1}c(1-x^2)dx=1$
$\Rightarrow 2c(1-\frac{1}{3})=1$
$\Rightarrow 2c\left(\frac{2}{3}\right)=1$
$\Rightarrow c=\frac{3}{4}$
$p=3,q=4$
$\therefore p+q=7$