What is the integral of (y^2)/(16) ?

The integral of (y^2)/(16) is (y^3)/(48)+C

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integral of (y^2)/(16)

\int \frac{y ^{2}}{16} d y

\frac{y ^{3}}{48} + C

Solution steps

\int \frac{y ^{2}}{16} d y

Take the constant out:

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

= \frac{1}{16} \cdot \int y^{2} d y

Apply the Power Rule

: \frac{y ^{3}}{3}

= \frac{1}{16} \cdot \frac{y ^{3}}{3}

Simplify

\frac{1}{16} \cdot \frac{y ^{3}}{3} : \frac{y ^{3}}{48}

= \frac{y ^{3}}{48}

Add a constant to the solution

= \frac{y ^{3}}{48} + C

y^{^{''}} - y = 0

n = 0 \sum \infty \frac{n !}{10 8 ^{n}}

\int \frac{sec ^{2} ( 5 x )}{tan ( 5 x ) - 4} d x

y^{^{'}} = 4 y + a

f (x) = 2 x - x