Question

Prove that the area common to the two parabolas y = 2x² and y = x² + 4 is $\frac{32}{3}$ sq. units.

Answer 1

We have, two parabolas y = 2x² and y = x² + 4

$$\begin{gathered} \mathrm{To}\mathrm{find}\mathrm{point}\mathrm{of}\mathrm{intersection},\mathrm{solve}\mathrm{the}\mathrm{two}\mathrm{equations} \\ 2x^2=x^2+4 \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \\ \therefore y=4 \\ \mathrm{Thus}A(2,4)\mathrm{and}A\text{'}(-2,4)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{parabolas} \\ \mathrm{Shaded}\mathrm{area}=2\times \mathrm{area}\left(OCAO\right) \\ \mathrm{Consider}\mathrm{a}\mathrm{vertical}\mathrm{stip}\mathrm{of}\mathrm{length}\left|y_2-y_1\right|\mathrm{and}\mathrm{width}dx \\ \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|y_2-y_1\right|dx \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0tox=2 \\ \mathrm{Area}\left(OCAO\right)={\int }_0^2\left|y_2-y_1\right|dx={\int }_0^2\left(y_2-y_1\right)dx\left\{\because \left|y_2-y_1\right|=y_2-y_1\mathrm{as}{y}_2>y_1\right\} \\ ={\int }_0^2\left(x^2+4-2x^2\right)dx \\ ={\int }_0^2\left(4-x^2\right)dx \\ ={\left[\left(4x-\frac{x^3}{3}\right)\right]}_0^2 \\ =8-\frac{8}{2} \\ =\frac{16}{3}\mathrm{sq}\mathrm{units} \\ \mathrm{Shaded}\mathrm{area}=2\times \mathrm{area}\left(OCAO\right)=2\times \frac{16}{3}=\frac{32}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$