integral from 0 to 1 of 2/(t^2+1)

Lösung

\int_{0}^{1} \frac{2}{t ^{2} + 1} d t

\frac{π}{2}

Dezimale

1.57079 \dots

Schritte zur Lösung

\int_{0}^{1} \frac{2}{t ^{2} + 1} d t

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

= 2 [arctan (t)]_{0}^{1}

Berechne die Grenzen:

\frac{π}{4}

= 2 \cdot \frac{π}{4}

Vereinfache

= \frac{π}{2}

d er i v a t i v e f (x) = 2 π r^{2} + \frac{1000}{r}

t an g e n t y = \frac{6}{x}, (36, \frac{6}{6})

\frac{d y}{d x} = - 2 y + e^{x}

\int \frac{2}{x} + 3 sin (x) - 4 e^{x} d x

\frac{\partial}{\partial x} (6 z e^{x yz})