integral from 0 to 1/2 of 2/(81+4x^2)

Lösung

\int_{0}^{\frac{1}{2}} \frac{2}{81 + 4 x ^{2}} d x

\frac{1}{9} arctan (\frac{1}{9})

Dezimale

0.01229 \dots

Schritte zur Lösung

\int_{0}^{\frac{1}{2}} \frac{2}{81 + 4 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}{2}} \frac{1}{81 + 4 x ^{2}} d x

Wende integrale Substitution an

= 2 \cdot \int_{0}^{\frac{1}{9}} \frac{1}{18 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{18} \cdot \int_{0}^{\frac{1}{9}} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= 2 \cdot \frac{1}{18} [arctan (u)]_{0}^{\frac{1}{9}}

Vereinfache

2 \cdot \frac{1}{18} [arctan (u)]_{0}^{\frac{1}{9}} : \frac{1}{9} [arctan (u)]_{0}^{\frac{1}{9}}

= \frac{1}{9} [arctan (u)]_{0}^{\frac{1}{9}}

Berechne die Grenzen:

arctan (\frac{1}{9})

= \frac{1}{9} arctan (\frac{1}{9})

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