integral from 2 to 5 of 4/(9+(x-2)^2)

Lösung

\int_{2}^{5} \frac{4}{9 + ( x - 2 ) ^{2}} d x

\frac{π}{3}

Dezimale

1.04719 \dots

Schritte zur Lösung

\int_{2}^{5} \frac{4}{9 + ( x - 2 ) ^{2}} d x

Entferne die Konstante:

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

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

Wende U-Substitution an

= 4 \cdot \int_{0}^{3} \frac{1}{9 + u ^{2}} d u

Wende integrale Substitution an

= 4 \cdot \int_{0}^{1} \frac{1}{3 ( v ^{2} + 1 )} d v

Entferne die Konstante:

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

= 4 \cdot \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{v ^{2} + 1} d v

Nutze das gemeinsame Integral :

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

= 4 \cdot \frac{1}{3} [arctan (v)]_{0}^{1}

Vereinfache

4 \cdot \frac{1}{3} [arctan (v)]_{0}^{1} : \frac{4}{3} [arctan (v)]_{0}^{1}

= \frac{4}{3} [arctan (v)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{4}{3} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{3}

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