What is the integral from 0 to 1 of (x(1+3y^2))/4 ?

The integral from 0 to 1 of (x(1+3y^2))/4 is x/2

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integral from 0 to 1 of (x(1+3y^2))/4

\int_{0}^{1} \frac{x ( 1 + 3 y ^{2} )}{4} d y

\frac{x}{2}

Solution steps

\int_{0}^{1} \frac{x ( 1 + 3 y ^{2} )}{4} d y

Take the constant out:

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

= \frac{x}{4} \cdot \int_{0}^{1} 1 + 3 y^{2} 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{x}{4} (\int_{0}^{1} 1 d y + \int_{0}^{1} 3 y^{2} d y)

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

\int_{0}^{1} 3 y^{2} d y = 1

= \frac{x}{4} (1 + 1)

Simplify

\frac{x}{4} (1 + 1) : \frac{x}{2}

= \frac{x}{2}

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What is the integral from 0 to 1 of (x(1+3y^2))/4 ?
The integral from 0 to 1 of (x(1+3y^2))/4 is x/2