How do you evaluate the definite integral $\int x^{2} + 2 x - 2$ $int x^2+2x-2$ from $[- 3, 0]$ $[-3,0]$ ?

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Answer 1 · 2017-03-10T16:12:22

The answer is $= - 6$ $=-6$

We apply

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx=x^(n+1)/(n+1)+C$

Therefore,

$\int_{- 3}^{0} (x^{2} + 2 x - 2) d x$ $int_-3^0(x^2+2x-2)dx$

$= {[\frac{x^{3}}{3} + \frac{2}{2} x^{2} - 2 x]}_{- 3}^{2}$ $=[x^3/3+2/2x^2-2x]_-3^0$

$= 0 - (- \frac{27}{3} + 9 + 6)$ $=0-(-27/3+9+6)$

$= - 6$ $=-6$

How do you evaluate the definite integral ∫x2+2x−2int x^2+2x-2 from [−3,0][-3,0]?