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

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

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

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

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

Solution steps

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

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

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

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

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

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

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