Question

Using integration, find the area of the region bounded by the line y – 1 = x, the x-axis and the ordinates x = – 2, x = 3

Answer 1

The equation of given line is y – 1 = x

y = x + 1

The line y = x + 1 meets the x-axis at x = – 1

Since x varies from – 2 to 3

Hence a part of lies below the x-axis and the other part lies above the x-axis.

∴ Area A = `int_(-2)^(-1) (- y)  "d"x + int_(-1)^3 y  "d"x`

= `int_(-2)^(-1) (x + 1)  "d"x + int_(-1)^3 (x + 1)  "d"x`

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

= `{[(-1)^2/2 + (-1)] - [(-2)^2/2 + (-2)]} + {[(3)^2/2 + (3)] - [(-1)^2/2 + (-1)]}`

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

= `- {(-1)/2 - 0} + 0 + {9/2 + 3 + 1/2}`

= `1/2 + 9/2 + 3 + 1/2`

= `(1 + 9 + 6 + 1)/2`

A = `17/2` sq.units