integral from 0 to 8/3 of 1/(9x^2+64)

Lösung

\int_{0}^{\frac{8}{3}} \frac{1}{9 x ^{2} + 64} d x

\frac{π}{96}

Dezimale

0.03272 \dots

Schritte zur Lösung

\int_{0}^{\frac{8}{3}} \frac{1}{9 x ^{2} + 64} d x

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{24 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{24} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{24} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{24} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{96}

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