integral from 0 to 4 of (4-t)(sqrt(t))

Lösung

\int_{0}^{4} (4 - t) (t) d t

\frac{128}{15}

Dezimale

8.53333 \dots

Schritte zur Lösung

\int_{0}^{4} (4 - t) t d t

Multipliziere aus

(4 - t) t : 4 t - t^{\frac{3}{2}}

= \int_{0}^{4} 4 t - t^{\frac{3}{2}} 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_{0}^{4} 4 t d t - \int_{0}^{4} t^{\frac{3}{2}} d t

\int_{0}^{4} 4 t d t = \frac{64}{3}

\int_{0}^{4} t^{\frac{3}{2}} d t = \frac{64}{5}

= \frac{64}{3} - \frac{64}{5}

Vereinfache

\frac{64}{3} - \frac{64}{5} : \frac{128}{15}

= \frac{128}{15}

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

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

\int_{4}^{8} 9 d x

\int_{0}^{2} x^{2} - (- x^{2} + 4 x) d x

\int_{- 8}^{7} (1 - x) d x