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

Lösung

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

\frac{4}{15}

Dezimale

0.26666 \dots

Schritte zur Lösung

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

Wende die Summenregel an:

\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}

Vereinfache

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

= \frac{4}{15}

\frac{d}{d x} (4 (e^{x})^{9})

in v erse l a pl a ce \frac{s + 2}{( s + 1 ) ( s - s ^{2} )}

\int \frac{8 x ^{2} - 4 x + 7}{4 x ^{3} + x ^{2} + 4 x + 1} d x

t an g e n t f (x) = - x^{2} - 4 x + 2, at x = - 1

\frac{d}{d x} (\frac{x ^{2} - 36}{x + 6})