integral from 0 to 1 of [3x+cos(3x-3)]

Lösung

\int_{0}^{1} [3 x + cos (3 x - 3)] d x

\frac{3}{2} + \frac{1}{3} sin (3)

Dezimale

1.54704 \dots

Schritte zur Lösung

\int_{0}^{1} 3 x + cos (3 x - 3) 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_{0}^{1} 3 x d x + \int_{0}^{1} cos (3 x - 3) d x

\int_{0}^{1} 3 x d x = \frac{3}{2}

\int_{0}^{1} cos (3 x - 3) d x = \frac{1}{3} sin (3)

= \frac{3}{2} + \frac{1}{3} sin (3)

\int_{0}^{r} r^{3} d r

\int_{0}^{x} \frac{2 t - 3}{4 t ^{2} + 7} d t

\int_{1}^{e} \frac{x ^{3} + 1}{1 + x ^{2}} d x

\int_{0}^{2} (1 - x) d x

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