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

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

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

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

4

Solution steps

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

Eliminate Absolutes

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

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

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

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

= \frac{4}{3} + \frac{4}{3} + \frac{4}{3}

Simplify

\frac{4}{3} + \frac{4}{3} + \frac{4}{3} : 4

= 4

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