What is the integral from-8 to 0 of (y/8+sqrt(y+9)) ?

The integral from-8 to 0 of (y/8+sqrt(y+9)) is 40/3

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integral from-8 to 0 of (y/8+sqrt(y+9))

\int_{- 8}^{0} (\frac{y}{8} + y + 9) d y

\frac{40}{3}

Solution steps

\int_{- 8}^{0} \frac{y}{8} + y + 9 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 8}^{0} \frac{y}{8} d y + \int_{- 8}^{0} y + 9 d y

\int_{- 8}^{0} \frac{y}{8} d y = - 4

\int_{- 8}^{0} y + 9 d y = \frac{52}{3}

= - 4 + \frac{52}{3}

Simplify

- 4 + \frac{52}{3} : \frac{40}{3}

= \frac{40}{3}

x \to \infty lim (x + 9)

\int \frac{1}{x ^{2} + 3 x + 3} d x

x \to 0 lim (- arcsin (x))

\int_{0}^{1} 9 cos (\frac{π t}{2}) d t

What is the integral from-8 to 0 of (y/8+sqrt(y+9)) ?
The integral from-8 to 0 of (y/8+sqrt(y+9)) is 40/3