So my friend forwarded me this question from his math class:
The parabolas $y = cx^2$ and $y = 1 - x^2$ intersect in the first quadrant as shown below. Find $c$ so that the areas of the two shaded regions are equal.
I'm supposed to be the math genius here, but I can't figure it out! My integrals are all over the place. Help me impress my friend :)
On
There is in fact no need to solve for the intersection point. Note that if the white curved "triangle" bounded by the two curves and the $x$-axis is added to either shaded region, we get the entire area under the associated curve.
So we just need to find $c$ such that
$$ \int_{x=0}^1 cx^2 \, dx = \int_{x=0}^1 1-x^2 \, dx $$
Can you take it from here?
You wish to solve
$$\int_0^{x_0} 1-(1+c)x^2 \ dx = \int_{x_0}^1 (c+1)x^2-1 \ dx$$
where
$$x_0=\dfrac{1}{\sqrt{c+1}}$$
which is found by solving $1-x^2=cx^2$ and noting $x_0>0$.