What is the integral from 1 to 3 of ((3x-y))/5 ?

The integral from 1 to 3 of ((3x-y))/5 is 1/5 (-4+6x)

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integral from 1 to 3 of ((3x-y))/5

\int_{1}^{3} \frac{( 3 x - y )}{5} d y

\frac{1}{5} (- 4 + 6 x)

Solution steps

\int_{1}^{3} \frac{3 x - y}{5} d y

Take the constant out:

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

= \frac{1}{5} \cdot \int_{1}^{3} - y + 3 x 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}{5} (- \int_{1}^{3} y d y + \int_{1}^{3} 3 x d y)

\int_{1}^{3} y d y = 4

\int_{1}^{3} 3 x d y = 6 x

= \frac{1}{5} (- 4 + 6 x)

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What is the integral from 1 to 3 of ((3x-y))/5 ?
The integral from 1 to 3 of ((3x-y))/5 is 1/5 (-4+6x)