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

Question

3996 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-22T17:41:55

The answer is #=32/3#

We need

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

#int_2^4(x^2-4)dx#

#=[x^3/3-4x]_2^4#

#=(64/3-16)-(8/3-8)#

#=64/3-8/3-16+8#

#=56/3-8#

#=32/3#