What is the integral from 0 to 4 of |sqrt(x+2)-x| ?

The integral from 0 to 4 of |sqrt(x+2)-x| is -4sqrt(6)+28/3+(16-4sqrt(2))/3

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integral from 0 to 4 of |sqrt(x+2)-x|

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

- 46 + \frac{28}{3} + \frac{16 - 4 2}{3}

Decimal Notation

2.98308 \dots

Solution steps

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

Eliminate Absolutes

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

\int_{0}^{2} x + 2 - x d x = \frac{16 - 4 2}{3} - 2

\int_{2}^{4} - x + 2 + x d x = - 46 + \frac{34}{3}

= \frac{16 - 4 2}{3} - 2 - 46 + \frac{34}{3}

Simplify

\frac{16 - 4 2}{3} - 2 - 46 + \frac{34}{3} : - 46 + \frac{28}{3} + \frac{16 - 4 2}{3}

= - 46 + \frac{28}{3} + \frac{16 - 4 2}{3}

\int \frac{y ^{2} + 16}{y} d y

\frac{d}{d x} (ln (\frac{x ^{3} - 2 x}{4 + x}))

f (x) = \frac{2 x + 1}{x + 2}

\frac{d}{d x} (\frac{x}{2})

11 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 9

What is the integral from 0 to 4 of |sqrt(x+2)-x| ?
The integral from 0 to 4 of |sqrt(x+2)-x| is -4sqrt(6)+28/3+(16-4sqrt(2))/3