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

Lösung

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

\frac{π}{18}

Dezimale

0.17453 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

\frac{1}{9 + 9 x ^{2}} = \frac{1}{9} \frac{1}{1 + x ^{2}}

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

= 6 \cdot \frac{1}{9} [arctan (x)]_{1}^{3}

Vereinfache

6 \cdot \frac{1}{9} [arctan (x)]_{1}^{3} : \frac{2}{3} [arctan (x)]_{1}^{3}

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

Berechne die Grenzen:

\frac{π}{12}

= \frac{2}{3} \cdot \frac{π}{12}

Vereinfache

= \frac{π}{18}

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