Sketch the region whose area is represented by the definite integral and eveluate the integral using an appropriate formula form geometry
(a) `int_1^(2)2dx`
(b) `int_-1^(2)(x+2)dx`
(c) `int_0^(2)(x-1)dx`

Explanation

(a). The graph of the integrand is the horizontal line `y=2`, sothe region is a rectangle of height 2 extending over the interval from 1 to 4.
`int_(1)^(4)2dx=`(area of rectangle)
`=2(3)=6`
(b). The graph of the integrand is the line `y=x+2`, so the region is a trapezoid whose base extends from `x=-1` to `x=2` [Fig. (b)]
Thus.
`int_(-1)^(2)(x+2)dx=`(area of trapezoid)`=(1)/(2)(1+4)(3)=(15)/(2)`
(c ). The graph of `y=x-1` is shown in figure and we leave it for you to verify that the shaded triangular regions both have area `(1)/(2)` over the interval `[0,2]` the net signed is `A_1-A_2=(1)/(2)-(1)/(2)=0`,
and over the interval `[0,1]` the net signed area is `-A_2=-(1)/(2)`,
Thus,
`int_(0)^(2)(x-1)dx=0` and `int_(0)^(1)(x-1)dx=-(1)/(2)`