integral from-1 to 2 of (x+1)

Lösung

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

\frac{9}{2}

Dezimale

4.5

Schritte zur Lösung

\int_{- 1}^{2} 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

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

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

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

= \frac{3}{2} + 3

Vereinfache

\frac{3}{2} + 3 : \frac{9}{2}

= \frac{9}{2}

\int_{1}^{3} 95 + 768 x^{2} - 16 x d x

\int_{- 0.1}^{0.5} sin (2 x) d x

\int_{0}^{\frac{π}{2}} \frac{1}{x} d x

\int_{e^{- 1}}^{e} ln (x) d x

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