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

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KollegeApply · Accepted Answer

$\frac{32}{3}$\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 - x^$\frac{3}{3}$ \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 } [Solution diagram: 44030_solution_1.png]

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