Question

Sketch the graph of y = |x + 3| and find the area of the region enclosed by the curve, x-axis, between x = − 6 and x = 0, using integration.

Answer 1

Graph of y = |x +3|

y = |x +3| = `{(x + 3, x +3 ≥ 0), (-(x + 3), x + 3 < 0):}`

 = `{(x + 3, "for"  x ≥ -3), (-(x + 3), "for"  x < -3):}`

Now required area A = `int_(-6)^0|x + 3|dx`

= `int_(-6)^(-3)-(x + 3)dx + int_(-3)^0(x + 3) dx`

= `-[x^2/2 + 3x]_(-6)^(-3) + [x^2/2 + 3x]_(-3)^0`

= `-[((-3)^2/2 + 3(-3)) - ((-6)^2/2 + 3(-6))] + [(0)^2/2 + 3(0)] - [((-3)^2/2 + 3(-3))]`

= `-[(9/2 - 9) - (36/2 - 18)] + [0 - (9/2 - 9)]`

= `-[((9 - 18)/2) - ((36 - 36)/2)] + [0 - ((9 - 18)/2)]`

= `-[(-9)/2 - 0] + 9/2`

= `9/2 + 9/2`

= `(9 + 9)/2`

= `18/2`

= 9 sq. unit