integral from 0 to 9/4 of 1/(16x^2+81)

Lösung

\int_{0}^{\frac{9}{4}} \frac{1}{16 x ^{2} + 81} d x

\frac{π}{144}

Dezimale

0.02181 \dots

Schritte zur Lösung

\int_{0}^{\frac{9}{4}} \frac{1}{16 x ^{2} + 81} d x

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{36 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{36} \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}{36} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{36} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{144}

\int_{0}^{1} x 4 - 3 x d x

\int_{0}^{4} \frac{1}{49 x ^{2} + 25} d x

\int_{1}^{10} \frac{3}{x} d x

\int_{1}^{2} \frac{z ^{2} + 5}{z} d z

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