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

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Answer 1 · 2017-05-07T16:55:16

#:. int_-1^2(-x^2+x+2)dx=9/2=4.5#

By the Fundamental Theorem of Calculus, we know that,

#intf(x)dx=F(x)+c rArr int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a)#

#:. int_-1^2(-x^2+x+2)dx=[-x^3/3+x^2/2+2x]_-1^2#

#=(-8/3+2+4)-{-(-1)/3+1/2-2}#

#=(-8/3-1/3)+6+2-1/2#

#=-3+8-1/2#

#:. int_-1^2(-x^2+x+2)dx=9/2=4.5#