How do you evaluate the definite integral #int (x^2-9)dx# from [3,4]?

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Answer 1 · 2016-11-13T13:44:30

The answer is #=10/3#

To evaluate the integral, we use

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

#int_3^4(x^2-9)dx= int_3^4x^2dx-int_3^4 9dx#

#=[x^3/3-9x ]_3^4#

#=(64/3-36)-(9-27)#

#=64/3-36-9+27#

#=10/3#