What is the integral from 0 to 1 of 36pi(x^2-x^{10}) ?

The integral from 0 to 1 of 36pi(x^2-x^{10}) is (96pi)/(11)

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integral from 0 to 1 of 36pi(x^2-x^{10})

\int_{0}^{1} 36 π (x^{2} - x^{10}) d x

\frac{96 π}{11}

Decimal Notation

27.41753 \dots

Solution steps

\int_{0}^{1} 36 π (x^{2} - x^{10}) d x

Take the constant out:

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

= 36 π \cdot \int_{0}^{1} x^{2} - x^{10} 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

= 36 π (\int_{0}^{1} x^{2} d x - \int_{0}^{1} x^{10} d x)

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

\int_{0}^{1} x^{10} d x = \frac{1}{11}

= 36 π (\frac{1}{3} - \frac{1}{11})

Simplify

36 π (\frac{1}{3} - \frac{1}{11}) : \frac{96 π}{11}

= \frac{96 π}{11}

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What is the integral from 0 to 1 of 36pi(x^2-x^{10}) ?
The integral from 0 to 1 of 36pi(x^2-x^{10}) is (96pi)/(11)