integral from 1 to 4 of (t^2-t-6)

Lösung

\int_{1}^{4} (t^{2} - t - 6) d t

- \frac{9}{2}

Dezimale

- 4.5

Schritte zur Lösung

\int_{1}^{4} t^{2} - t - 6 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

= \int_{1}^{4} t^{2} d t - \int_{1}^{4} t d t - \int_{1}^{4} 6 d t

\int_{1}^{4} t^{2} d t = 21

\int_{1}^{4} t d t = \frac{15}{2}

\int_{1}^{4} 6 d t = 18

= 21 - \frac{15}{2} - 18

Vereinfache

21 - \frac{15}{2} - 18 : - \frac{9}{2}

= - \frac{9}{2}

\int_{0}^{t} 2 t d t

\int_{0}^{3} \frac{2 x}{( x ^{2} + 2 ) ^{3}} d x

\int_{1}^{3} \frac{5}{t} d t

\int_{0}^{4} x - \frac{x ^{2}}{16} d x

\int_{0}^{4} \frac{1}{4 x - x ^{2}} d x