Question

The area common to the parabola y = 2x² and y = x² + 4 is
(a) $\frac{2}{3}$sq. units

(b) $\frac{3}{2}$sq. units

(c) $\frac{32}{3}$sq. units

(d) $\frac{3}{32}$sq. units

Answer 1

$\left(\mathrm{c}\right)\frac{32}{3}\mathrm{sq}.\mathrm{units}$



Common region of two given parabola y = 2x² and y = x² + 4 is infinite as we see in the figure here.
Therefore, area common to these two parabola is infinity.

DISCLAIMER:
In the question, instead of
"The area common to the parabola y = 2x² and y = x² + 4 is"
It should be
"The closed area made by the parabola y = 2x² and y = x² + 4 is"
Solution of this question is as follow.



To find the point of intersection of the parabolas equate the equations y = 2x² and y = x² + 4 we get

$$\begin{gathered} 2x^2=x^2+4 \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \\ \therefore y=8 \end{gathered}$$

Therefore, the points of intersection are A(−2, 8) and C(2, 8).

Therefore, the required area ABCD,
$$\begin{gathered} A={\int }_{-2}^2\left(y_1-y_2\right)\mathit{d}x\mathit{}\left(\mathrm{Where},y_1=x^2+4\mathrm{and}{y}_2=2x^2\right) \\ ={\int }_{-2}^2\left(x^2+4-2x^2\right)\mathit{}\mathit{d}x \\ ={\int }_{-2}^2\left(4-x^2\right)\mathit{d}x \\ ={\left[4x-\frac{x^3}{3}\right]}_{-2}^2 \\ =\left[4\left(2\right)-\frac{{\left(2\right)}^3}{3}\right]-\left[4\left(-2\right)-\frac{{\left(-2\right)}^3}{3}\right] \\ =\left[8-\frac{8}{3}\right]-\left[-8+\frac{8}{3}\right] \\ =8-\frac{8}{3}+8-\frac{8}{3} \\ =16-\frac{16}{3} \\ =\frac{32}{3}\mathrm{square}\mathrm{units} \end{gathered}$$