What is the integral from 0 to 1 of pi(y-y^2) ?

The integral from 0 to 1 of pi(y-y^2) is pi/6

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integral from 0 to 1 of pi(y-y^2)

\int_{0}^{1} π (y - y^{2}) d y

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{1} π (y - y^{2}) d y

Take the constant out:

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

= π \cdot \int_{0}^{1} y - 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

= π (\int_{0}^{1} y d y - \int_{0}^{1} y^{2} d y)

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

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

= π (\frac{1}{2} - \frac{1}{3})

Simplify

π (\frac{1}{2} - \frac{1}{3}) : \frac{π}{6}

= \frac{π}{6}

\frac{\partial}{\partial x} (x e^{z})

\int e^{i t} d t

x \to \infty lim (\frac{7}{x} - 1)

y^{^{''}} + 5 y^{^{'}} + 4 y = sin (t)

What is the integral from 0 to 1 of pi(y-y^2) ?
The integral from 0 to 1 of pi(y-y^2) is pi/6