What is the integral from 0 to 9 of |sqrt(2x+8)-x| ?

The integral from 0 to 9 of |sqrt(2x+8)-x| is 49/2+(128-16sqrt(2)-26sqrt(26))/3

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integral from 0 to 9 of |sqrt(2x+8)-x|

\int_{0}^{9} 2 x + 8 - x d x

\frac{49}{2} + \frac{128 - 16 2 - 26 26}{3}

Decimal

15.43269 \dots

Solution steps

\int_{0}^{9} 2 x + 8 - x d x

Eliminate Absolutes

= \int_{0}^{4} 2 x + 8 - x d x + \int_{4}^{9} - 2 x + 8 + x d x

\int_{0}^{4} 2 x + 8 - x d x = \frac{16 ( 4 - 2 )}{3} - 8

\int_{4}^{9} - 2 x + 8 + x d x = - \frac{2 ( 13 26 - 32 )}{3} + \frac{65}{2}

= \frac{16 ( 4 - 2 )}{3} - 8 - \frac{2 ( 13 26 - 32 )}{3} + \frac{65}{2}

Simplify

\frac{16 ( 4 - 2 )}{3} - 8 - \frac{2 ( 13 26 - 32 )}{3} + \frac{65}{2} : \frac{49}{2} + \frac{128 - 16 2 - 26 26}{3}

= \frac{49}{2} + \frac{128 - 16 2 - 26 26}{3}

\int_{0}^{\frac{π}{4}} 1 - 4 sin^{2} (x) d x

\int 2 u sin (u) d u

\frac{d y}{d x} = \frac{x + y}{x}

x \to 5 lim (x^{5} - 9 x + 1)

\int - 4 sin (x) d x

What is the integral from 0 to 9 of |sqrt(2x+8)-x| ?
The integral from 0 to 9 of |sqrt(2x+8)-x| is 49/2+(128-16sqrt(2)-26sqrt(26))/3