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

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

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

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

$=-2-2+9/2-3$

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

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