What is the integral from 0 to 9 of |sqrt(x)-2| ?

The integral from 0 to 9 of |sqrt(x)-2| is 16/3

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integral from 0 to 9 of |sqrt(x)-2|

\int_{0}^{9} x - 2 d x

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

\int_{0}^{9} x - 2 d x

Eliminate Absolutes

= \int_{0}^{4} - x + 2 d x + \int_{4}^{9} x - 2 d x

\int_{0}^{4} - x + 2 d x = \frac{8}{3}

\int_{4}^{9} x - 2 d x = \frac{8}{3}

= \frac{8}{3} + \frac{8}{3}

Simplify

\frac{8}{3} + \frac{8}{3} : \frac{16}{3}

= \frac{16}{3}

What is the integral from 0 to 9 of |sqrt(x)-2| ?
The integral from 0 to 9 of |sqrt(x)-2| is 16/3