integral from 0 to 4 of 1/(9x^2+25)

Lösung

\int_{0}^{4} \frac{1}{9 x ^{2} + 25} d x

\frac{1}{15} arctan (\frac{12}{5})

Dezimale

0.07840 \dots

Schritte zur Lösung

\int_{0}^{4} \frac{1}{9 x ^{2} + 25} d x

Wende integrale Substitution an

= \int_{0}^{\frac{12}{5}} \frac{1}{15 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{15} \cdot \int_{0}^{\frac{12}{5}} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{15} [arctan (u)]_{0}^{\frac{12}{5}}

Berechne die Grenzen:

arctan (\frac{12}{5})

= \frac{1}{15} arctan (\frac{12}{5})

\int_{\frac{1}{2}}^{1} \frac{1}{x ^{2}} e^{\frac{1}{x}} d x

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

\int_{- 2}^{2} \frac{x ^{2}}{2} - 3 d x

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

\int_{0}^{0.5} 1 - \frac{x}{2} d x