integral from 1 to 3 of (2x-1/x)^2

Lösung

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

\frac{82}{3}

Dezimale

27.33333 \dots

Schritte zur Lösung

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

Multipliziere aus

(2 x - \frac{1}{x})^{2} : 4 x^{2} - 4 + \frac{1}{x ^{2}}

= \int_{1}^{3} 4 x^{2} - 4 + \frac{1}{x ^{2}} 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}^{3} 4 x^{2} d x - \int_{1}^{3} 4 d x + \int_{1}^{3} \frac{1}{x ^{2}} d x

\int_{1}^{3} 4 x^{2} d x = \frac{104}{3}

\int_{1}^{3} 4 d x = 8

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

= \frac{104}{3} - 8 + \frac{2}{3}

Vereinfache

\frac{104}{3} - 8 + \frac{2}{3} : \frac{82}{3}

= \frac{82}{3}

(\frac{6}{x})^{^{'}}

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

x \to \infty lim ((5)^{x})

\frac{\partial}{\partial x} (e^{s i n (x)} cos (x))

f (x) = 7 x^{- 6} - 5 x^{- 3}