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

The integral from 0 to 3y of (3+x^2+1/4 y^2) is 9y+9y^3+(3y^3)/4

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

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

9 y + 9 y^{3} + \frac{3 y ^{3}}{4}

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{3 y} 3 d x = 9 y

\int_{0}^{3 y} x^{2} d x = 9 y^{3}

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

= 9 y + 9 y^{3} + \frac{3 y ^{3}}{4}

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What is the integral from 0 to 3y of (3+x^2+1/4 y^2) ?
The integral from 0 to 3y of (3+x^2+1/4 y^2) is 9y+9y^3+(3y^3)/4