How do you evaluate $int x^2 + x + 4dx$ for [0, 2]?

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Answer 1 · 2015-09-21T07:45:54

$38/3$

$int_0^2x^2+x+4dx=int_0^2x^2dx+int_0^2xdx+int_0^24dx =[x^3/3]_0^2+[x^2/2]_0^2+4[x]_0^2=8/3+4/2+8-0=38/3$

How do you evaluate int x^2 + x + 4dxint x^2 + x + 4dx for [0, 2]?