integral from 0 to 2 of x^3sqrt(1+9x^4)

Lösung

\int_{0}^{2} x^{3} 1 + 9 x^{4} d x

\frac{145 145 - 1}{54}

Dezimale

32.31539 \dots

Schritte zur Lösung

\int_{0}^{2} x^{3} 1 + 9 x^{4} d x

Wende U-Substitution an

= \int_{1}^{145} \frac{u}{36} d u

Entferne die Konstante:

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

= \frac{1}{36} \cdot \int_{1}^{145} u d u

Wende die Potenzregel an

= \frac{1}{36} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{145}

Berechne die Grenzen:

\frac{290 145}{3} - \frac{2}{3}

= \frac{1}{36} (\frac{290 145}{3} - \frac{2}{3})

Vereinfache

= \frac{145 145 - 1}{54}

\frac{d}{d x} (8 arctan (x))

\int_{1}^{e} \frac{1}{x} d x

d er i v a t i v e y = ln (\frac{1}{x x + 1})

x y^{^{'}} - 2 y - 16 = 0

y^{^{'}} = e^{5 y} sin (6 x)