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

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

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

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

\frac{4}{15}

Decimal Notation

0.26666 \dots

Solution steps

\int_{- 1}^{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

= \int_{- 1}^{1} x^{2} d x - \int_{- 1}^{1} x^{4} d x

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

\int_{- 1}^{1} x^{4} d x = \frac{2}{5}

= \frac{2}{3} - \frac{2}{5}

Simplify

\frac{2}{3} - \frac{2}{5} : \frac{4}{15}

= \frac{4}{15}

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