What is the integral from 0 to-8 of (x+1)^2 ?

The integral from 0 to-8 of (x+1)^2 is -344/3

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integral from 0 to-8 of (x+1)^2

\int_{0}^{- 8} (x + 1)^{2} d x

- \frac{344}{3}

Decimal

- 114.66666 \dots

Solution steps

\int_{0}^{- 8} (x + 1)^{2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 8}^{0} (x + 1)^{2} d x

Expand

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

= - \int_{- 8}^{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_{- 8}^{0} x^{2} d x + \int_{- 8}^{0} 2 x d x + \int_{- 8}^{0} 1 d x)

\int_{- 8}^{0} x^{2} d x = \frac{512}{3}

\int_{- 8}^{0} 2 x d x = - 64

\int_{- 8}^{0} 1 d x = 8

= - (\frac{512}{3} - 64 + 8)

Simplify

- (\frac{512}{3} - 64 + 8) : - \frac{344}{3}

= - \frac{344}{3}

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What is the integral from 0 to-8 of (x+1)^2 ?
The integral from 0 to-8 of (x+1)^2 is -344/3