integral from 0 to 3 of 1/(x^2+16)

Lösung

\int_{0}^{3} \frac{1}{x ^{2} + 16} d x

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

Dezimale

0.16087 \dots

Schritte zur Lösung

\int_{0}^{3} \frac{1}{x ^{2} + 16} d x

Wende integrale Substitution an

= \int_{0}^{\frac{3}{4}} \frac{1}{4 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{\frac{3}{4}} \frac{1}{u ^{2} + 1} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{4} [arctan (u)]_{0}^{\frac{3}{4}}

Berechne die Grenzen:

arctan (\frac{3}{4})

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

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

\int_{\frac{π}{2}}^{π} sin^{4} (x) d x

\int_{0}^{5} \frac{w}{w ^{2} - 2} d w

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

\int_{0}^{2 - x} \frac{3}{4} y d y