How do you evaluate the definite integral #int (2+x)dx# from [0,4]?

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Answer 1 · 2017-01-18T08:48:41

16

#int_0^4(2+x)dx#

#intax^ndx=ax^(n+1)/(n+1)+C; n!=-1#

we have

#int_0^4(2+x)dx=[2x+x^2/2]_0^4#

no need for the constant with limits

#[2x+x^2/2]^4-cancel([2x+x^2/2]_0)#

#(2xx4+4^2/2)=8+8=16#