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

The integral from 1 to 3 of (x^2-4x+6) is 14/3

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integral from 1 to 3 of (x^2-4x+6)

\int_{1}^{3} (x^{2} - 4 x + 6) d x

\frac{14}{3}

Decimal Notation

4.66666 \dots

Solution steps

\int_{1}^{3} 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}^{3} x^{2} d x - \int_{1}^{3} 4 x d x + \int_{1}^{3} 6 d x

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

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

\int_{1}^{3} 6 d x = 12

= \frac{26}{3} - 16 + 12

Simplify

\frac{26}{3} - 16 + 12 : \frac{14}{3}

= \frac{14}{3}

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What is the integral from 1 to 3 of (x^2-4x+6) ?
The integral from 1 to 3 of (x^2-4x+6) is 14/3