What is the integral from 0 to 1 of (x+y)/z ?

The integral from 0 to 1 of (x+y)/z is 1/z (y+1/2)

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integral from 0 to 1 of (x+y)/z

\int_{0}^{1} \frac{x + y}{z} d x

\frac{1}{z} (y + \frac{1}{2})

Solution steps

\int_{0}^{1} \frac{x + y}{z} d x

Take the constant out:

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

= \frac{1}{z} \cdot \int_{0}^{1} y + x d x

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_{0}^{1} y d x + \int_{0}^{1} x d x)

\int_{0}^{1} y d x = y

\int_{0}^{1} x d x = \frac{1}{2}

= \frac{1}{z} (y + \frac{1}{2})

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\int_{0}^{2} 1 + e^{2 x} d x

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What is the integral from 0 to 1 of (x+y)/z ?
The integral from 0 to 1 of (x+y)/z is 1/z (y+1/2)