What is the integral from 1 to 8 of (x^2-4x+6) ?

The integral from 1 to 8 of (x^2-4x+6) is 259/3

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integral from 1 to 8 of (x^2-4x+6)

\int_{1}^{8} (x^{2} - 4 x + 6) d x

\frac{259}{3}

Decimal Notation

86.33333 \dots

Solution steps

\int_{1}^{8} x^{2} - 4 x + 6 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} 4 x d x + \int_{1}^{8} 6 d x

\int_{1}^{8} x^{2} d x = \frac{511}{3}

\int_{1}^{8} 4 x d x = 126

\int_{1}^{8} 6 d x = 42

= \frac{511}{3} - 126 + 42

Simplify

\frac{511}{3} - 126 + 42 : \frac{259}{3}

= \frac{259}{3}

f (x) = ln (x - x^{2})

\int 2 x^{2} - 3 x d x

\frac{2 - x}{8}

\int x \cdot e^{3 x} d x

\frac{d ^{2} y}{d x ^{2}} + 9 y = 10 cos (2 x)

What is the integral from 1 to 8 of (x^2-4x+6) ?
The integral from 1 to 8 of (x^2-4x+6) is 259/3