What is the integral from 1 to 4 of |x^2-x-6| ?

The integral from 1 to 4 of |x^2-x-6| is 61/6

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integral from 1 to 4 of |x^2-x-6|

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

\frac{61}{6}

Decimal

10.16666 \dots

Solution steps

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

Eliminate Absolutes

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

\int_{1}^{3} - x^{2} + x + 6 d x = \frac{22}{3}

\int_{3}^{4} x^{2} - x - 6 d x = \frac{17}{6}

= \frac{22}{3} + \frac{17}{6}

Simplify

\frac{22}{3} + \frac{17}{6} : \frac{61}{6}

= \frac{61}{6}

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What is the integral from 1 to 4 of |x^2-x-6| ?
The integral from 1 to 4 of |x^2-x-6| is 61/6