Question

The area enclosed between the curves y = 2x² and y = 6 is
1. \(2 \sqrt3\)
2. \(4 \sqrt3\)
3. \(6 \sqrt3\)
4. \(8 \sqrt3\)

Answer 1

Correct Answer - Option 4 : \(8 \sqrt3\)

Concept:

The area enclosed between the 2 curves y₁ = f(x) and y₂ = g(x)

A = \(\rm \left|\int \limits_{x_1}^{x_2}f(x) -g(x)dx\right|\)

Here x₁ and x₂ are the limits of the area to be found out.

Calculation:

At y = 6, 

6 = 2x²

x = ± √3

f(x) = y₁ = 2x²

g(x) = y₂ = 6

Area under the curve A = \(\rm\left|\int \limits_{-\sqrt3}^{\sqrt3}(2x^2 - 6) dx\right|\)

A = \(\rm\left|\left[{2x^3\over3} - 6x \right]_{-\sqrt3}^{\sqrt3}\right|\)

A = \(\rm\left|\left[{2({\sqrt3})^3\over3} - 6{\sqrt3} \right]-\left[{2({-\sqrt3})^3\over3} - 6(-{\sqrt3})\right]\right|\)

A = \(\rm\left|\left[- 4{\sqrt3} \right]-\left[4{\sqrt3}\right]\right|\)

A = \(\rm 8\sqrt3\)