How do you evaluate the definite integral #int x^2+2x-2# from #[-3,0]#?

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Answer 1 · 2017-03-10T16:12:22

The answer is #=-6#

We apply

#intx^ndx=x^(n+1)/(n+1)+C#

Therefore,

#int_-3^0(x^2+2x-2)dx#

#=[x^3/3+2/2x^2-2x]_-3^0#

#=0-(-27/3+9+6)#

#=-6#