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

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Answer 1 · 2016-11-15T07:34:32

The answer is #=16/3#

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

#intx^3dx=x^4/4+C#

#intxdx=x^2/2 +C#

Therefore #int_0^2(x^3/3+2x)dx#

#=[x^4/12+2x^2/2]_0^2#

#=[x^4/12+x^2]_0^2#

#=(16/12+4)#

#=64/12=16/3#