What is the integral from 0 to 4 of |x^2-3x+2| ?

The integral from 0 to 4 of |x^2-3x+2| is 17/3

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integral from 0 to 4 of |x^2-3x+2|

\int_{0}^{4} x^{2} - 3 x + 2 d x

\frac{17}{3}

Decimal Notation

5.66666 \dots

Solution steps

\int_{0}^{4} x^{2} - 3 x + 2 d x

Eliminate Absolutes

= \int_{0}^{1} x^{2} - 3 x + 2 d x + \int_{1}^{2} - x^{2} + 3 x - 2 d x + \int_{2}^{4} x^{2} - 3 x + 2 d x

\int_{0}^{1} x^{2} - 3 x + 2 d x = \frac{5}{6}

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

\int_{2}^{4} x^{2} - 3 x + 2 d x = \frac{14}{3}

= \frac{5}{6} + \frac{1}{6} + \frac{14}{3}

Simplify

\frac{5}{6} + \frac{1}{6} + \frac{14}{3} : \frac{17}{3}

= \frac{17}{3}

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What is the integral from 0 to 4 of |x^2-3x+2| ?
The integral from 0 to 4 of |x^2-3x+2| is 17/3