integral from 2 to 4 of (2+sqrt(u))^2

Lösung

\int_{2}^{4} (2 + u)^{2} d u

\frac{4 ( 16 - 4 2 )}{3} + 14

Dezimale

27.79086 \dots

Schritte zur Lösung

\int_{2}^{4} (2 + u)^{2} d u

Multipliziere aus

(2 + u)^{2} : 4 + 4 u + u

= \int_{2}^{4} 4 + 4 u + u d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{4} 4 d u + \int_{2}^{4} 4 u d u + \int_{2}^{4} u d u

\int_{2}^{4} 4 d u = 8

\int_{2}^{4} 4 u d u = \frac{4 ( 16 - 4 2 )}{3}

\int_{2}^{4} u d u = 6

= 8 + \frac{4 ( 16 - 4 2 )}{3} + 6

Vereinfache

= \frac{4 ( 16 - 4 2 )}{3} + 14

\frac{d}{d x} (- ln (x) cos (x))

\int (5 cos (3 x) - 4 sin (5 x)) d x

n = 5 \sum \infty \frac{5 ^{n}}{1 2 ^{n}}

x \to \infty lim (\frac{9}{x} - 1)

d er i v a t i v e f (x) = \frac{6 x ^{5} + 4 x}{x ^{4} + 3}