How do you evaluate the definite integral #int(-x-1)dx# from #[-3,2]#?

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Answer 1 · 2016-08-17T16:35:21

#-5/2, or, -2.5#.

Let #I=int_-3^2 (-x-1)dx#.

#:. I=-int_-3^2 (x+1)dx#

#=-[x^2/2+x]_-3^2#

#=-(2^2/2+2)+{(-3)^2/2+(-3)}#

#=-2-2+9/2-3#

#=-5/2, or, -2.5#.