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

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Answer 1 · 2017-01-21T07:41:46

The answer is #=e^2-e+5/2#

We need

#inte^xdx=e^x+C#

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

Therefore,

#int_1^2(e^x+x+1)dx#

#=[e^x+x^2/2+x]_1^2#

#=(e^2+2+2)-(e+1/2+1)#

#=e^2-e+4-3/2#

#=e^2-e+5/2#