What is the integral from 2 to 4 of (-x^2+6x-8) ?

The integral from 2 to 4 of (-x^2+6x-8) is 4/3

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integral from 2 to 4 of (-x^2+6x-8)

\int_{2}^{4} (- x^{2} + 6 x - 8) d x

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

\int_{2}^{4} - x^{2} + 6 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_{2}^{4} x^{2} d x + \int_{2}^{4} 6 x d x - \int_{2}^{4} 8 d x

\int_{2}^{4} x^{2} d x = \frac{56}{3}

\int_{2}^{4} 6 x d x = 36

\int_{2}^{4} 8 d x = 16

= - \frac{56}{3} + 36 - 16

Simplify

- \frac{56}{3} + 36 - 16 : \frac{4}{3}

= \frac{4}{3}

\frac{d}{d x} (x^{3} (8 x^{2} + 2 x - 3)^{2})

x \to + 0 lim (\frac{2 x + 1}{x ^{3}})

\frac{\partial}{\partial y} (x y e^{- 3 y})

x^{2} - 3 x, 5 x, [0, 8]

x^{2} - 1, 0, 2

What is the integral from 2 to 4 of (-x^2+6x-8) ?
The integral from 2 to 4 of (-x^2+6x-8) is 4/3