integral from 0 to 6 of pi(sqrt(x-1))^2

Lösung

\int_{0}^{6} π (x - 1)^{2} d x

12 π

Dezimale

37.69911 \dots

Schritte zur Lösung

\int_{0}^{6} π (x - 1)^{2} d x

Entferne die Konstante:

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

= π \cdot \int_{0}^{6} (x - 1)^{2} d x

Vereinfache

(x - 1)^{2} : x - 1

= π \cdot \int_{0}^{6} 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

= π (\int_{0}^{6} x d x - \int_{0}^{6} 1 d x)

\int_{0}^{6} x d x = 18

\int_{0}^{6} 1 d x = 6

= π (18 - 6)

Vereinfache

= 12 π

x^{2} y^{^{'}} + 8 x y = 17 y^{3}

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

\frac{d}{d x} (1 + ln (x))

i = 1 \sum \infty 3 (\frac{1}{4})^{i - 1}

x \to 1 lim (\frac{in ( x )}{x - 1})