integral from 1 to 5 of x/(sqrt(2x-1))

Lösung

\int_{1}^{5} \frac{x}{2 x - 1} d x

\frac{16}{3}

Dezimale

5.33333 \dots

Schritte zur Lösung

\int_{1}^{5} \frac{x}{2 x - 1} d x

Wende U-Substitution an

= \int_{1}^{3} \frac{u ^{2} + 1}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{1}^{3} u^{2} + 1 d u

Wende die Summenregel an:

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

= \frac{1}{2} (\int_{1}^{3} u^{2} d u + \int_{1}^{3} 1 d u)

\int_{1}^{3} u^{2} d u = \frac{26}{3}

\int_{1}^{3} 1 d u = 2

= \frac{1}{2} (\frac{26}{3} + 2)

Vereinfache

\frac{1}{2} (\frac{26}{3} + 2) : \frac{16}{3}

= \frac{16}{3}

x \to 0 lim (cos (x) + sin (x))

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{x ^{2} - 1}, y (7) = 7

y^{^{'}} = 8 x^{3} + 2 \frac{y}{x}

\frac{d y}{d t} = - 2 y, y (0) = 2

n = 1 \sum \infty n (\frac{5}{4})^{n}