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

The integral from 0 to 1 of ((y^2+y^3))/2 is 7/24

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

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

\frac{7}{24}

Decimal Notation

0.29166 \dots

Solution steps

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

Take the constant out:

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

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

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

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

= \frac{1}{2} (\frac{1}{3} + \frac{1}{4})

Simplify

\frac{1}{2} (\frac{1}{3} + \frac{1}{4}) : \frac{7}{24}

= \frac{7}{24}

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What is the integral from 0 to 1 of ((y^2+y^3))/2 ?
The integral from 0 to 1 of ((y^2+y^3))/2 is 7/24