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

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

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

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

\frac{1 + 3 y ^{2}}{2}

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{2}

= \frac{1 + 3 y ^{2}}{4} [\frac{x ^{2}}{2}]_{0}^{2}

Compute the boundaries:

2

= \frac{1 + 3 y ^{2}}{4} \cdot 2

Simplify

= \frac{1 + 3 y ^{2}}{2}

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