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 32/3. (Enter your answers as a comma-separated list.)

Answer 1

Answer:

-2,2

Step-by-step explanation:

Let

$y_1=4x^2-c^2$

$y_2=c^2-4x^2$

We have to find the value of c such that the are of the region bounded by the parabolas =32/3

$y_1=y_2$

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

$4x^2+4x^2=c^2+c^2$

$8x^2=2c^2$

$x^2=c^2/4$

$x=\pm \frac{c}{2}$

Now, the area bounded by two curves

$A=\int_{a}^{b}(y_2-y_1)dx$

$A=\int_{-c/2}^{c/2}(c^2-4x^2-4x^2+c^2)dx$

$\frac{32}{3}=\int_{-c/2}^{c/2}(2c^2-8x^2)dx$

$\frac{32}{3}=2\int_{-c/2}^{c/2}(c^2-4x^2)dx$

$\frac{32}{3}=2[c^2x-\frac{4}{3}x^3]^{c/2}_{-c/2}$

$\frac{32}{3}=2(c^2(c/2+c/2)-4/3(c^3/8+c^3/28))$

$\frac{32}{3}=2(c^3-\frac{4}{3}(\frac{c^3}{4}))$

$\frac{32}{3}=2(c^3-\frac{c^3}{3})$

$\frac{32}{3}=2(\frac{2}{3}c^3)$

$c^3=\frac{32\times 3}{4\times 3}$

$c^3=8$

$c=\sqrt[3]{8}=2$

When c=2 and when c=-2 then the given parabolas gives the same answer.

Therefore, value of c=-2, 2