integral from 1 to 2 of sqrt(x-1)-x+1

Lösung

\int_{1}^{2} x - 1 - x + 1 d x

\frac{1}{6}

Dezimale

0.16666 \dots

Schritte zur Lösung

\int_{1}^{2} x - 1 - x + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} x - 1 d x - \int_{1}^{2} x d x + \int_{1}^{2} 1 d x

\int_{1}^{2} x - 1 d x = \frac{2}{3}

\int_{1}^{2} x d x = \frac{3}{2}

\int_{1}^{2} 1 d x = 1

= \frac{2}{3} - \frac{3}{2} + 1

Vereinfache

\frac{2}{3} - \frac{3}{2} + 1 : \frac{1}{6}

= \frac{1}{6}

t an g e n t f (x) = x^{s i n (x)}, (\frac{π}{2}, \frac{π}{2})

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

a re a y = x, y = (x - 2)^{2}

x \to 0 lim (\frac{2 x}{x + 4})

\frac{d}{d x} (\frac{1 - tan ( x )}{1 + tan ( x )})