integral from 0 to 1/6 of 3/(sqrt(1-9x^2))

Lösung

\int_{0}^{\frac{1}{6}} \frac{3}{1 - 9 x ^{2}} d x

\frac{π}{6}

Dezimale

0.52359 \dots

Schritte zur Lösung

\int_{0}^{\frac{1}{6}} \frac{3}{1 - 9 x ^{2}} d x

Entferne die Konstante:

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

= 3 \cdot \int_{0}^{\frac{1}{6}} \frac{1}{1 - 9 x ^{2}} d x

Trigonometrische Substitution anwenden

= 3 \cdot \int_{0}^{\frac{π}{6}} \frac{1}{3} d u

Entferne die Konstante:

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

= 3 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{π}{6}} 1 d u

Integral einer Konstanten :

\int a d x = a x

= 3 \cdot \frac{1}{3} [1 \cdot u]_{0}^{\frac{π}{6}}

Vereinfache

3 \cdot \frac{1}{3} [1 \cdot u]_{0}^{\frac{π}{6}} : [u]_{0}^{\frac{π}{6}}

= [u]_{0}^{\frac{π}{6}}

Berechne die Grenzen:

\frac{π}{6}

= \frac{π}{6}

\frac{d x}{d t} = x t

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 8, y^{^{'}} (0) = 2

\int - 8 x^{2} d x

in v erse l a pl a ce \frac{e ^{- π (s)}}{s ^{3} + s}

\frac{d}{d x} ((x + \frac{1}{x})^{5})