integral from 0 to 5 of (2e^x+4cos(x))

Lösung

\int_{0}^{5} (2 e^{x} + 4 cos (x)) d x

2 (e^{5} - 1) + 4 sin (5)

Dezimale

290.99062 \dots

Schritte zur Lösung

\int_{0}^{5} 2 e^{x} + 4 cos (x) 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}^{5} 2 e^{x} d x + \int_{0}^{5} 4 cos (x) d x

\int_{0}^{5} 2 e^{x} d x = 2 (e^{5} - 1)

\int_{0}^{5} 4 cos (x) d x = 4 sin (5)

= 2 (e^{5} - 1) + 4 sin (5)

x \to 2 - lim (\frac{3 x}{x ^{2} - 4})

\frac{d}{d x} (\frac{e ^{x}}{3 x ^{2} + 2 x ^{4}})

\frac{\partial}{\partial x} (2 x sin (8 x^{2} y))

y^{^{'}} = 0.0012 y (1700 - y)

y^{^{'}} = \frac{( 2 x ^{2} + 3 y ^{2} )}{2 x y}