Answer:
-2,2
Step-by-step explanation:
Let
[tex]y_1=4x^2-c^2[/tex]
[tex]y_2=c^2-4x^2[/tex]
We have to find the value of c such that the are of the region bounded by the parabolas =32/3
[tex]y_1=y_2[/tex]
[tex]4x^2-c^2=c^2-4x^2[/tex]
[tex]4x^2+4x^2=c^2+c^2[/tex]
[tex]8x^2=2c^2[/tex]
[tex]x^2=c^2/4[/tex]
[tex]x=\pm \frac{c}{2}[/tex]
Now, the area bounded by two curves
[tex]A=\int_{a}^{b}(y_2-y_1)dx[/tex]
[tex]A=\int_{-c/2}^{c/2}(c^2-4x^2-4x^2+c^2)dx[/tex]
[tex]\frac{32}{3}=\int_{-c/2}^{c/2}(2c^2-8x^2)dx[/tex]
[tex]\frac{32}{3}=2\int_{-c/2}^{c/2}(c^2-4x^2)dx[/tex]
[tex]\frac{32}{3}=2[c^2x-\frac{4}{3}x^3]^{c/2}_{-c/2}[/tex]
[tex]\frac{32}{3}=2(c^2(c/2+c/2)-4/3(c^3/8+c^3/28))[/tex]
[tex]\frac{32}{3}=2(c^3-\frac{4}{3}(\frac{c^3}{4}))[/tex]
[tex]\frac{32}{3}=2(c^3-\frac{c^3}{3})[/tex]
[tex]\frac{32}{3}=2(\frac{2}{3}c^3)[/tex]
[tex]c^3=\frac{32\times 3}{4\times 3}[/tex]
[tex]c^3=8[/tex]
[tex]c=\sqrt[3]{8}=2[/tex]
When c=2 and when c=-2 then the given parabolas gives the same answer.
Therefore, value of c=-2, 2
