integral from 0 to u of v+x

Lösung

\int_{0}^{u} v + x d v

ux + \frac{u ^{2}}{2}

Schritte zur Lösung

\int_{0}^{u} v + x d v

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}^{u} x d v + \int_{0}^{u} v d v

\int_{0}^{u} x d v = ux

\int_{0}^{u} v d v = \frac{u ^{2}}{2}

= ux + \frac{u ^{2}}{2}

(1 - t^{2}) y^{^{'}} - t^{2} y = (1 + t) \cdot (1 - t^{2})

t an g e n t 3 x^{2} + y^{2} = 1, (\frac{1}{2}, \frac{- 1}{2})

y^{^{''}} - y^{^{'}} + 12 y = 0

\frac{d}{d x} (\frac{( e ^{x} )}{1 - e ^{- x}})

\int - \frac{x}{x ^{2} + y ^{2}} d x