What is the integral from 0 to 1 of pi(x^4-x^2) ?

The integral from 0 to 1 of pi(x^4-x^2) is -pi 2/15

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integral from 0 to 1 of pi(x^4-x^2)

\int_{0}^{1} π (x^{4} - x^{2}) d x

- π \frac{2}{15}

Decimal

- 0.41887 \dots

Solution steps

\int_{0}^{1} π (x^{4} - x^{2}) d x

Take the constant out:

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

= π \cdot \int_{0}^{1} x^{4} - x^{2} 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

= π (\int_{0}^{1} x^{4} d x - \int_{0}^{1} x^{2} d x)

\int_{0}^{1} x^{4} d x = \frac{1}{5}

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

= π (\frac{1}{5} - \frac{1}{3})

Simplify

π (\frac{1}{5} - \frac{1}{3}) : - π \frac{2}{15}

= - π \frac{2}{15}

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\int_{5}^{10} \frac{18}{25} x d x

What is the integral from 0 to 1 of pi(x^4-x^2) ?
The integral from 0 to 1 of pi(x^4-x^2) is -pi 2/15