integral from 2 to x of (3-t)/2

Lösung

\int_{2}^{x} \frac{3 - t}{2} d t

\frac{6 x - x ^{2} - 8}{4}

Schritte zur Lösung

\int_{2}^{x} \frac{3 - t}{2} d t

Entferne die Konstante:

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

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

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_{2}^{x} 3 d t - \int_{2}^{x} t d t)

\int_{2}^{x} 3 d t = 3 x - 6

\int_{2}^{x} t d t = \frac{x ^{2}}{2} - 2

= \frac{1}{2} (3 x - 6 - (\frac{x ^{2}}{2} - 2))

Vereinfache

\frac{1}{2} (3 x - 6 - (\frac{x ^{2}}{2} - 2)) : \frac{6 x - x ^{2} - 8}{4}

= \frac{6 x - x ^{2} - 8}{4}

\int_{- 2}^{x} 100 - t^{4} d t

\int_{1}^{6} - 3 cos (\frac{nπ t}{4}) d t

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

\int_{0}^{2} (9 - x) x d x

\int_{1}^{8} \frac{2}{t} d t