integral from 0 to 7 of 1/(49+x^2)

Lösung

\int_{0}^{7} \frac{1}{49 + x ^{2}} d x

\frac{π}{28}

Dezimale

0.11219 \dots

Schritte zur Lösung

\int_{0}^{7} \frac{1}{49 + x ^{2}} d x

Wende integrale Substitution an

= \int_{0}^{1} \frac{1}{7 ( u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{7} \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}{7} [arctan (u)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= \frac{1}{7} \cdot \frac{π}{4}

Vereinfache

= \frac{π}{28}

\frac{\partial}{\partial x} (3 y^{5} x^{2} + 8 y x^{3})

x \to 3 - lim (\frac{7 - x - 4}{x - 3})

(1 + x^{14}) (\frac{d y}{d x}) + 14 x^{13} y = 11 x^{5}

t an g e n t 3^{x^{2} - 1}, at x = 2

\int 3^{5 x - 2} d x