integral from 2 to 1 of (1+z/2)

Lösung

\int_{2}^{1} (1 + \frac{z}{2}) d z

- \frac{7}{4}

Dezimale

- 1.75

Schritte zur Lösung

\int_{2}^{1} 1 + \frac{z}{2} d z

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{2} 1 + \frac{z}{2} d z

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} 1 d z + \int_{1}^{2} \frac{z}{2} d z)

\int_{1}^{2} 1 d z = 1

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

= - (1 + \frac{3}{4})

Füge

1 + \frac{3}{4}

zusammen:

\frac{7}{4}

= - \frac{7}{4}

\int_{1}^{x} sin (x) d x

\int_{0}^{1} e^{x} - e^{- 2 x} d x

\int_{0}^{2} x e^{\frac{- x ^{2} - 1}{2}} d x

\int_{1}^{4} - x - (2 - x) d x

\int_{1}^{4} (2 x) d x