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

Lösung

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

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

Dezimale

0.43520 \dots

Schritte zur Lösung

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

Wende integrale Substitution an

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

Nutze das gemeinsame Integral :

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

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

Berechne die Grenzen:

arctan (\frac{1}{2})

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

Vereinfache

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

\int_{- 1}^{1} 3 x + 1 d x

\int_{1}^{3} \frac{6}{9 + 9 x ^{2}} d x

\int_{4}^{6} 5 x d x

\int_{2}^{18} (4 x^{3} - 2 x + 2) d x

\int_{0}^{2} (x^{3} - x) d x