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

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

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

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

\frac{2}{15}

Decimal Notation

0.13333 \dots

Solution steps

\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

= \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}{3} - \frac{1}{5}

Simplify

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

= \frac{2}{15}

\frac{d}{d x} (\frac{3 x}{4 x ^{2} + 9})

x^{2} - 17, 16 x

y^{^{'}} = 12 - \frac{y}{30}

\frac{d}{d x} (\frac{2 x}{sin ( x )})

\frac{b ( - 12 b - 290 )}{8 - b}

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