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

Lösung

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

\frac{π}{36}

Dezimale

0.08726 \dots

Schritte zur Lösung

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

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{9 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{9} \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}{9} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{9} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{36}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 20 e^{t}, y (0) = - 4, y^{^{'}} (0) = 1

\int_{1}^{4} 6 e^{x} d x

cr i t i c a l f (x, y) = (x)^{2} + (y)^{4}

x \to 3 + lim (\frac{1}{∣ x - 3 ∣})

y \to 0 lim (\frac{y ^{2}}{y ^{4}})