Question

The closed area made by the parabola y = 2x² and y = x² + 4 is __________ .

Options

• \[\frac{2}{3}\]sq. units

• \[\frac{3}{2}\]sq. units

• \[\frac{32}{3}\]sq. units

• \[\frac{3}{32}\]sq. units

Answer 1

\[\frac{32}{3}\text{ sq . units }\]

 

To find the point of intersection of the parabolas equate the equations y = 2x² and y = x² + 4 we get 
\[2 x^2 = x^2 + 4\]
\[ \Rightarrow x^2 = 4\]
\[ \Rightarrow x = \pm 2\]
\[ \therefore y = 8\]
Therefore, the points of intersection are A(−2, 8) and C(2, 8).
Therefore, the required area ABCD,
\[A = \int_{- 2}^2 \left( y_1 - y_2 \right) d x ...........\left(\text{Where }, y_1 = x^2 + 4\text{ and }y_2 = 2 x^2 \right)\]
\[ = \int_{- 2}^2 \left( x^2 + 4 - 2 x^2 \right) d x\]
\[ = \int_{- 2}^2 \left( 4 - x^2 \right) d x\]
\[ = \left[ 4x - \frac{x^3}{3} \right]_{- 2}^2 \]
\[ = \left[ 4\left( 2 \right) - \frac{\left( 2 \right)^3}{3} \right] - \left[ 4\left( - 2 \right) - \frac{\left( - 2 \right)^3}{3} \right]\]
\[ = \left[ 8 - \frac{8}{3} \right] - \left[ - 8 + \frac{8}{3} \right]\]
\[ = 8 - \frac{8}{3} + 8 - \frac{8}{3}\]
\[ = 16 - \frac{16}{3}\]
\[ = \frac{32}{3}\text{ square units }\]