What is the integral of (1-y^2)/2 ?

The integral of (1-y^2)/2 is 1/2 (y-(y^3)/3)+C

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integral of (1-y^2)/2

\int \frac{1 - y ^{2}}{2} d y

\frac{1}{2} (y - \frac{y ^{3}}{3}) + C

Solution steps

\int \frac{1 - y ^{2}}{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 1 - 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{1}{2} (\int 1 d y - \int y^{2} d y)

\int 1 d y = y

\int y^{2} d y = \frac{y ^{3}}{3}

= \frac{1}{2} (y - \frac{y ^{3}}{3})

Add a constant to the solution

= \frac{1}{2} (y - \frac{y ^{3}}{3}) + C

What is the integral of (1-y^2)/2 ?
The integral of (1-y^2)/2 is 1/2 (y-(y^3)/3)+C