What is the integral from-2 to 0 of (x+1)^2-1 ?

The integral from-2 to 0 of (x+1)^2-1 is -4/3

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integral from-2 to 0 of (x+1)^2-1

\int_{- 2}^{0} (x + 1)^{2} - 1 d x

- \frac{4}{3}

Decimal

- 1.33333 \dots

Solution steps

\int_{- 2}^{0} (x + 1)^{2} - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 2}^{0} (x + 1)^{2} d x - \int_{- 2}^{0} 1 d x

\int_{- 2}^{0} (x + 1)^{2} d x = \frac{2}{3}

\int_{- 2}^{0} 1 d x = 2

= \frac{2}{3} - 2

Simplify

\frac{2}{3} - 2 : - \frac{4}{3}

= - \frac{4}{3}

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What is the integral from-2 to 0 of (x+1)^2-1 ?
The integral from-2 to 0 of (x+1)^2-1 is -4/3