What is the integral from 3x to 8 of (x+y)/(60) ?

The integral from 3x to 8 of (x+y)/(60) is (-15x^2+16x+64)/(120)

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integral from 3x to 8 of (x+y)/(60)

\int_{3 x}^{8} \frac{x + y}{60} d y

\frac{- 15 x ^{2} + 16 x + 64}{120}

Solution steps

\int_{3 x}^{8} \frac{x + y}{60} d y

Take the constant out:

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

= \frac{1}{60} \cdot \int_{3 x}^{8} x + 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}{60} (\int_{3 x}^{8} x d y + \int_{3 x}^{8} y d y)

\int_{3 x}^{8} x d y = x \cdot 8 - 3 x^{2}

\int_{3 x}^{8} y d y = 32 - \frac{9 x ^{2}}{2}

= \frac{1}{60} (x \cdot 8 - 3 x^{2} + 32 - \frac{9 x ^{2}}{2})

Simplify

\frac{1}{60} (x \cdot 8 - 3 x^{2} + 32 - \frac{9 x ^{2}}{2}) : \frac{- 15 x ^{2} + 16 x + 64}{120}

= \frac{- 15 x ^{2} + 16 x + 64}{120}

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What is the integral from 3x to 8 of (x+y)/(60) ?
The integral from 3x to 8 of (x+y)/(60) is (-15x^2+16x+64)/(120)