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

Lösung

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

\frac{1}{2} arctan (2)

Dezimale

0.55357 \dots

Schritte zur Lösung

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Berechne die Grenzen:

arctan (2)

= \frac{1}{2} arctan (2)

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