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.
So far, i've tried to find the intersection point of the two functions, and I got that they intersect when $x = \sqrt{\frac{1}{c+1}}$. I also tried to use this info to get the area of the first and second regions and got $\int_0^\sqrt{\frac{1}{c+1}} (1-x^2)-(cx^2) dx$ and $\int_\sqrt{\frac{1}{c+1}}^1 (cx^2)-(1-x^2) dx$. I got stuck here, and don't know how to continue from there. also I want to know if I'm doing this right.
Let $b= 1/\sqrt{c+1}$ then your integral is $$ \int_0^b (1-x^2)-(cx^2) dx = \int_0^b(1-(1+c)x^2)dx $$ and $$\int_b^1 (cx^2)-(1-x^2) dx = \int_b^1((1+c)x^2-1)dx . $$ Then perform the integrals, insert the limits, and work out what value you need for $c$ to make the two regions have equal areas.