integral from 0 to 1/4 of 2/(1+16x^2)

Lösung

\int_{0}^{\frac{1}{4}} \frac{2}{1 + 16 x ^{2}} d x

\frac{π}{8}

Dezimale

0.39269 \dots

Schritte zur Lösung

\int_{0}^{\frac{1}{4}} \frac{2}{1 + 16 x ^{2}} d x

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{\frac{1}{4}} \frac{1}{1 + 16 x ^{2}} d x

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

= 2 \cdot \frac{1}{4} [arctan (u)]_{0}^{1}

Vereinfache

2 \cdot \frac{1}{4} [arctan (u)]_{0}^{1} : \frac{1}{2} [arctan (u)]_{0}^{1}

= \frac{1}{2} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{8}

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