integral from 1 to 3 of 1/(x^2+4)

Lösung

\int_{1}^{3} \frac{1}{x ^{2} + 4} d x

\frac{1}{2} (arctan (\frac{3}{2}) - arctan (\frac{1}{2}))

Dezimale

0.25957 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{1}{x ^{2} + 4} d x

Wende integrale Substitution an

= \int_{\frac{1}{2}}^{\frac{3}{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_{\frac{1}{2}}^{\frac{3}{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)]_{\frac{1}{2}}^{\frac{3}{2}}

Berechne die Grenzen:

arctan (\frac{3}{2}) - arctan (\frac{1}{2})

= \frac{1}{2} (arctan (\frac{3}{2}) - arctan (\frac{1}{2}))

\int_{\frac{1}{3}}^{3} (2 - \frac{2}{x ^{2}}) d x

\int_{0}^{4} 1000 x d x

\int_{1}^{2} \frac{1}{t ^{2} + 1} d t

\int_{0}^{π} cos (x) sin^{3} (x) d x

\int_{\frac{1}{4}}^{1} \frac{32}{x ^{3}} d x