integral from 0 to 1 of 5/(7+7x^2)

Lösung

\int_{0}^{1} \frac{5}{7 + 7 x ^{2}} d x

\frac{5 π}{28}

Dezimale

0.56099 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{5}{7 + 7 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}{7 + 7 x ^{2}} d x

\frac{1}{7 + 7 x ^{2}} = \frac{1}{7} \frac{1}{1 + x ^{2}}

= 5 \cdot \int_{0}^{1} \frac{1}{7} \cdot \frac{1}{1 + x ^{2}} d x

Entferne die Konstante:

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

= 5 \cdot \frac{1}{7} \cdot \int_{0}^{1} \frac{1}{1 + x ^{2}} d x

Nutze das gemeinsame Integral :

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

= 5 \cdot \frac{1}{7} [arctan (x)]_{0}^{1}

Vereinfache

5 \cdot \frac{1}{7} [arctan (x)]_{0}^{1} : \frac{5}{7} [arctan (x)]_{0}^{1}

= \frac{5}{7} [arctan (x)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{5}{7} \cdot \frac{π}{4}

Vereinfache

= \frac{5 π}{28}

\int_{0}^{4} (1 - x) d x

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