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

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

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

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

- \frac{1}{3}

Decimal

- 0.33333 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

(x + 1)^{2} : x^{2} + 2 x + 1

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

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

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

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

= - (\frac{1}{3} - 1 + 1)

Simplify

= - \frac{1}{3}

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