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

Lösung

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

\frac{4 2}{3} - 2

Dezimale

- 0.11438 \dots

Schritte zur Lösung

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

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

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

= \frac{4 2}{3} - 2

\int_{- 3}^{2} ∣ x + 1 ∣ d x

x \to 1 lim (\frac{2 x}{ln ( x )})

d er i v a t i v e f (x) = 6 - x

\int_{0}^{2} e^{\frac{x}{2}} d x

\frac{d y}{d x} = \frac{x y ^{3}}{1 + x ^{2}}