Find the area enclosed by the curve y = –x^2 and the straight line x + y + 2 = 0.

Question

Find the area enclosed by the curve y = –x² and the straight lilne x + y + 2 = 0.

Answer 1

y = -x²⇒ x² = -y

x² = -y parabola is not defined for positive values of y hence parabola will be below X-axis opening downwards and passing through (0, 0)

x + y + 2 = 0 is a straight line

To find point of intersection of parabola and straight line solve the parabola equation and the straight line equation simultaneously

Put y = -(x + 2) in x² = -y

⇒ x² = -(-(x + 2))

⇒ x² = x + 2

⇒ x² – x – 2 = 0

⇒ x² – 2x + x – 2 = 0

⇒ x(x – 2) + 1(x – 2) = 0

⇒ (x + 1)(x – 2) = 0

⇒ x = -1 and x = 2

Put x = -1 in x² = -y

⇒ (-1)² = -y

⇒ y = -1

Put x = 2 in x² = -y

⇒ 2² = -y

⇒ y = -4

Hence the parabola and line intersects at (-1, -1) and (2, -4)

Plot the parabola and straight line and the bounded area is as shown

⇒ area enclosed by line and parabola = area under line – area under parabola …(i)

Let us find area under the straight line

x + y + 2 = 0

⇒ y = -(x + 2)

Integrate from -1 to 2

Now let us find area under parabola

x² = -y

⇒ y = -x²

Integrate from -1 to 2

Using (i)

⇒ area enclosed by line and parabola = -15/2 -(-36) = -9/2 unit²

We are getting negative sign because the area is below X-axis as seen in figure

Hence area enclosed is 9/2 unit²