integral from 0 to 2 of 1/(9x^2+36)

Lösung

\int_{0}^{2} \frac{1}{9 x ^{2} + 36} d x

\frac{π}{72}

Dezimale

0.04363 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{1}{9 x ^{2} + 36} d x

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{18 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{18} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{18} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{18} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{72}

\int_{2}^{5} 2 x - 1 d x

\int_{0}^{1} \frac{3 y ^{2}}{4 + y ^{3}} d y

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

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

\int_{2}^{x^{2}} t^{2} t^{2} + 9 d t