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

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

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

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

\frac{1}{15}

Decimal Notation

0.06666 \dots

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} x^{2} - x^{4} 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

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

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

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

= \frac{1}{2} (\frac{1}{3} - \frac{1}{5})

Simplify

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

= \frac{1}{15}

\int_{- \infty}^{\infty} 2 d x

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\int_{0}^{3} x^{2} + 1 d x

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