integral from 1 to 2 of 2/(9+4x^2)

Lösung

\int_{1}^{2} \frac{2}{9 + 4 x ^{2}} d x

\frac{1}{3} (arctan (\frac{4}{3}) - arctan (\frac{2}{3}))

Dezimale

0.11309 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{2}{9 + 4 x ^{2}} d x

Entferne die Konstante:

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

= 2 \cdot \int_{1}^{2} \frac{1}{9 + 4 x ^{2}} d x

Wende integrale Substitution an

= 2 \cdot \int_{\frac{2}{3}}^{\frac{4}{3}} \frac{1}{6 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{6} \cdot \int_{\frac{2}{3}}^{\frac{4}{3}} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= 2 \cdot \frac{1}{6} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

Vereinfache

2 \cdot \frac{1}{6} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}} : \frac{1}{3} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

= \frac{1}{3} [arctan (u)]_{\frac{2}{3}}^{\frac{4}{3}}

Berechne die Grenzen:

arctan (\frac{4}{3}) - arctan (\frac{2}{3})

= \frac{1}{3} (arctan (\frac{4}{3}) - arctan (\frac{2}{3}))

\int_{1}^{3} 8 d x

\int_{0}^{\frac{π}{2}} cos (2 x - \frac{π}{2}) d x

\int_{0}^{1} r d r

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

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