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

y = x + 2 is a straight line

plot these equations and we have to find the area enclosed between them

To find the intersection points solve the equations x² = y and y = x + 2 simultaneously

Put y = x + 2 in x² = y

⇒ 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 and x = 2 in x² = y we get y = 1 and y = 4 respectively

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

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

Let us find area under line y = x + 2

⇒ y = x + 2

Integrate from -1 to 2

Now let us find area under the parabola

x² = y

⇒ y = x²

Integrate from -1 to 2

Using (i)

⇒ area enclosed by line and parabola = – 3 = unit²

Hence area enclosed is unit²