What is the integral from-2 to-1 of 1/z (y+1/2) ?

The integral from-2 to-1 of 1/z (y+1/2) is -1/z

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integral from-2 to-1 of 1/z (y+1/2)

\int_{- 2}^{- 1} \frac{1}{z} (y + \frac{1}{2}) d y

- \frac{1}{z}

Solution steps

\int_{- 2}^{- 1} \frac{1}{z} (y + \frac{1}{2}) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{z} \cdot \int_{- 2}^{- 1} \frac{1}{2} + y 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

= \frac{1}{z} (\int_{- 2}^{- 1} \frac{1}{2} d y + \int_{- 2}^{- 1} y d y)

\int_{- 2}^{- 1} \frac{1}{2} d y = \frac{1}{2}

\int_{- 2}^{- 1} y d y = - \frac{3}{2}

= \frac{1}{z} (\frac{1}{2} - \frac{3}{2})

Simplify

\frac{1}{z} (\frac{1}{2} - \frac{3}{2}) : - \frac{1}{z}

= - \frac{1}{z}

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What is the integral from-2 to-1 of 1/z (y+1/2) ?
The integral from-2 to-1 of 1/z (y+1/2) is -1/z