Question

The area enclosed between the two parabolas y=7-2x^2 and y=x^2+4 is

Answer 1

We have to find the area enclosed between the two parabolas y = 7 - 2x² and y = x² + 4

solution : find the points of intersection of two curves.

⇒7 - 2x² = x² + 4

⇒3 = 3x²

⇒x = -1, 1

application : The area between two curves, f(x) and g(x) is defined as $\int\limits^{x_1}_{x_0}|f(x)-g(x)|dx$,

where x₁ and x₀ are abscissa of the points of intersection.

area enclosed between the curves = $\int\limits^1_{-1}|7-2x^2-(x^2+4)|dx$

= $\int\limits^1_{-1}|3-3x^2|dx$

= $[3x-x^3]^1_{-1}$

= (3 × 1 - 1) - (3 × -1 + 1)

= 2 - (-2)

= 4

Therefore area enclosed between the two parabolas y = 7 - 2x² and y = x² + 4, is 4.