What is the integral from 0 to 1 of 6/5 (10x+4y) ?

The integral from 0 to 1 of 6/5 (10x+4y) is 6/5 (10x+2)

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integral from 0 to 1 of 6/5 (10x+4y)

\int_{0}^{1} \frac{6}{5} (10 x + 4 y) d y

\frac{6}{5} (10 x + 2)

Solution steps

\int_{0}^{1} \frac{6}{5} (10 x + 4 y) d y

Take the constant out:

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

= \frac{6}{5} \cdot \int_{0}^{1} 10 x + 4 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{6}{5} (\int_{0}^{1} 10 x d y + \int_{0}^{1} 4 y d y)

\int_{0}^{1} 10 x d y = 10 x

\int_{0}^{1} 4 y d y = 2

= \frac{6}{5} (10 x + 2)

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What is the integral from 0 to 1 of 6/5 (10x+4y) ?
The integral from 0 to 1 of 6/5 (10x+4y) is 6/5 (10x+2)