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

Lösung

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

\frac{ln ( 3 )}{6}

Dezimale

0.18310 \dots

Schritte zur Lösung

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{3} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{2}}

Vereinfache

\frac{1}{3} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{2}} : \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{2}}

= \frac{1}{3} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{2}}

Berechne die Grenzen:

\frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

= \frac{1}{3} \cdot \frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

Vereinfache

= \frac{ln ( 3 )}{6}

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