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

The integral from sqrt(8) to-1 of x^2-8 is -(16sqrt(2)+1)/3+16sqrt(2)+8

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

\int_{8}^{- 1} x^{2} - 8 d x

- \frac{16 2 + 1}{3} + 162 + 8

Decimal Notation

22.75161 \dots

Solution steps

\int_{8}^{- 1} x^{2} - 8 d x

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

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

\int_{- 1}^{8} x^{2} d x = \frac{16 2 + 1}{3}

\int_{- 1}^{8} 8 d x = 162 + 8

= - (\frac{16 2 + 1}{3} - (162 + 8))

Simplify

= - \frac{16 2 + 1}{3} + 162 + 8

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What is the integral from sqrt(8) to-1 of x^2-8 ?
The integral from sqrt(8) to-1 of x^2-8 is -(16sqrt(2)+1)/3+16sqrt(2)+8