Question

Find the values of c such that the area of the region bounded by the parabolas y = 4x2 − c2 and y = c2 − 4x2 is 256/3. (Enter your answers as a comma-separated list.) c = Incorrect: Your answer is incorrect.

Answer 1

The two parabolas intersect at $x$ such that

$4x^2-c^2=c^2-4x^2$

$\implies8x^2=2c^2\implies4x^2=c^2\implies x=\pm\dfrac c2$

Assume $c>0$. Then the area between the two curves is

$\displaystyle\int_{-c/2}^{c/2}\left|(4x^2-c^2)-(c^2-4x^2)\right|\,\mathrm dx=2\int_{-c/2}^{c/2}\left|4x^2-c^2\right|\,\mathrm dx$

The integrand is even, so the integral over this interval is equal to twice the integral over the positive half of the interval:

$\displaystyle4\int_0^{c/2}\left|4x^2-c^2\right|\,\mathrm dx$

Suppose $x=\dfrac c4$. Then

$4\left(\dfrac c4\right)^2-c^2=\dfrac{4c^2}{16}-c^2=-\dfrac34c^2$

which means $\left|4x^2-c^2\right|=c^2-4x^2$ by definition of absolute value.

We want the integral to have a value of 256/3:

$\displaystyle4\int_0^{c/2}(c^2-4x^2)\,\mathrm dx=\frac{256}3$

$\left(c^2x-\dfrac43x^3\right)\bigg|_0^{c/2}=\dfrac{64}3$

$c^2\left(\dfrac c2\right)-\dfrac43\left(\dfrac c2\right)^3=\dfrac{64}3$

$\dfrac{c^3}3=\dfrac{64}3$

$c^3=64\implies\boxed{c=4}$