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

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

Popular >

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

\int_{0}^{1} 2 π (x) (x - x^{2}) d x

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{1} 2 π x (x - x^{2}) d x

Take the constant out:

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

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

Expand

x (x - x^{2}) : x^{2} - x^{3}

= 2 π \cdot \int_{0}^{1} x^{2} - x^{3} 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

= 2 π (\int_{0}^{1} x^{2} d x - \int_{0}^{1} x^{3} d x)

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

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

= 2 π (\frac{1}{3} - \frac{1}{4})

Simplify

2 π (\frac{1}{3} - \frac{1}{4}) : \frac{π}{6}

= \frac{π}{6}

\frac{d}{d x} (lo g_{7} (\frac{x ^{4}}{5 x + 11}))

x^{2} (1 - 4 x)

\frac{d}{d x} (x (x^{3} + 1)^{\frac{1}{3}})

\frac{d}{d x} (x a^{2} + x^{2})

t^{3} \cdot sin (3 t)

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