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

Lösung

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

\frac{π}{6}

Dezimale

0.52359 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende integrale Substitution an

= 2 \cdot \int_{0}^{1} \frac{1}{3 ( 1 + u ^{2} )} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

Nutze das gemeinsame Integral :

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 2 \cdot \frac{1}{3} [arctan (u)]_{0}^{1}

Vereinfache

2 \cdot \frac{1}{3} [arctan (u)]_{0}^{1} : \frac{2}{3} [arctan (u)]_{0}^{1}

= \frac{2}{3} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{2}{3} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{6}

\int_{0}^{2} \frac{1}{x ^{2} - 5 x + 4} d x

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

x \to \frac{π}{3} lim (cos (x))

x \to 1 lim (\frac{- x + 1}{x - 1})

d er i v a t i v e x^{18}