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

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Answer 1 · 2017-01-18T08:48:41

16

$\int_{0}^{4} (2 + x) d x$ $int_0^4(2+x)dx$

$\int a x^{n} d x = a \frac{x^{n + 1}}{n + 1} + C; n \neq - 1$ $intax^ndx=ax^(n+1)/(n+1)+C; n!=-1$

we have

$\int_{0}^{4} (2 + x) d x = {[2 x + \frac{x^{2}}{2}]}_{0}^{4}$ $int_0^4(2+x)dx=[2x+x^2/2]_0^4$

no need for the constant with limits

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

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