What is the integral from-3 to 0 of (y/3+sqrt(y+4)) ?

The integral from-3 to 0 of (y/3+sqrt(y+4)) is 19/6

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integral from-3 to 0 of (y/3+sqrt(y+4))

\int_{- 3}^{0} (\frac{y}{3} + y + 4) d y

\frac{19}{6}

Decimal Notation

3.16666 \dots

Solution steps

\int_{- 3}^{0} \frac{y}{3} + y + 4 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_{- 3}^{0} \frac{y}{3} d y + \int_{- 3}^{0} y + 4 d y

\int_{- 3}^{0} \frac{y}{3} d y = - \frac{3}{2}

\int_{- 3}^{0} y + 4 d y = \frac{14}{3}

= - \frac{3}{2} + \frac{14}{3}

Simplify

- \frac{3}{2} + \frac{14}{3} : \frac{19}{6}

= \frac{19}{6}

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What is the integral from-3 to 0 of (y/3+sqrt(y+4)) ?
The integral from-3 to 0 of (y/3+sqrt(y+4)) is 19/6