What is the integral from-3 to-1 of (x^2-4x+8) ?

The integral from-3 to-1 of (x^2-4x+8) is 122/3

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integral from-3 to-1 of (x^2-4x+8)

\int_{- 3}^{- 1} (x^{2} - 4 x + 8) d x

\frac{122}{3}

Decimal Notation

40.66666 \dots

Solution steps

\int_{- 3}^{- 1} x^{2} - 4 x + 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_{- 3}^{- 1} x^{2} d x - \int_{- 3}^{- 1} 4 x d x + \int_{- 3}^{- 1} 8 d x

\int_{- 3}^{- 1} x^{2} d x = \frac{26}{3}

\int_{- 3}^{- 1} 4 x d x = - 16

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

= \frac{26}{3} - (- 16) + 16

Simplify

\frac{26}{3} - (- 16) + 16 : \frac{122}{3}

= \frac{122}{3}

x \to - 1 lim (3^{2 x})

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 4})

\frac{d}{d x} (\frac{2}{x ^{2} - 3})

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{x}

y = 34 (3 p + 1) - \frac{2}{5}, p > 0

What is the integral from-3 to-1 of (x^2-4x+8) ?
The integral from-3 to-1 of (x^2-4x+8) is 122/3