integral from 0 to 1 of 5/(sqrt(9-4x^2))

Lösung

\int_{0}^{1} \frac{5}{9 - 4 x ^{2}} d x

\frac{5}{2} arcsin (\frac{2}{3})

Dezimale

1.82431 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{5}{9 - 4 x ^{2}} d x

Entferne die Konstante:

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

= 5 \cdot \int_{0}^{1} \frac{1}{9 - 4 x ^{2}} d x

Trigonometrische Substitution anwenden

= 5 \cdot \int_{0}^{a r c s i n (\frac{2}{3})} \frac{1}{2} d u

Entferne die Konstante:

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

= 5 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{2}{3})} 1 d u

Integral einer Konstanten :

\int a d x = a x

= 5 \cdot \frac{1}{2} [1 \cdot u]_{0}^{a r c s i n (\frac{2}{3})}

Vereinfache

5 \cdot \frac{1}{2} [1 \cdot u]_{0}^{a r c s i n (\frac{2}{3})} : \frac{5}{2} [u]_{0}^{a r c s i n (\frac{2}{3})}

= \frac{5}{2} [u]_{0}^{a r c s i n (\frac{2}{3})}

Berechne die Grenzen:

arcsin (\frac{2}{3})

= \frac{5}{2} arcsin (\frac{2}{3})

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