integral from-1 to 0 of x/(sqrt(x+1))

Lösung

\int_{- 1}^{0} \frac{x}{x + 1} d x

- \frac{4}{3}

Dezimale

- 1.33333 \dots

Schritte zur Lösung

\int_{- 1}^{0} \frac{x}{x + 1} d x

Wende U-Substitution an

= \int_{0}^{1} 2 (u^{2} - 1) d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} 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

= 2 (\int_{0}^{1} u^{2} d u - \int_{0}^{1} 1 d u)

\int_{0}^{1} u^{2} d u = \frac{1}{3}

\int_{0}^{1} 1 d u = 1

= 2 (\frac{1}{3} - 1)

Vereinfache

2 (\frac{1}{3} - 1) : - \frac{4}{3}

= - \frac{4}{3}

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

\int_{2}^{5} \frac{6 - x}{x ^{3}} d x

\int_{0}^{1} e^{x} cos (x) d x

\int_{10}^{\infty} \frac{1}{x ^{6}} d x

\int_{\frac{π}{2}}^{\frac{5 π}{2}} sin (x) d x