Question

Find the area of the region bounded by curve 4x² = y and the line y = 8x + 12, using integration.

Answer 1

Given curve is 4x² = y and line is y = 8x + 12

On solving both equations, we get

4x² = 8x + 12

⇒ x² = 2x + 3

⇒ x² – 2x – 3 = 0

⇒ x = 3, –1

Required area = `int_-1^3 {(8x + 12) - 4x^2}dx`

= `4int_-1^3 (2x + 3 - x^2)dx`

= `4[x^2 + 3x - x^3/3]_-1^3`

= `4[(9 + 9 - 9) - (1 - 3 + 1/3)]`

= `4(9 + 2 - 1/3)`

= `4(11 - 1/3)`

= `4 xx 32/2`

= `128/2` sq.units