integral from y to y+1 of (x+y)

Lösung

\int_{y}^{y + 1} (x + y) d x

y + \frac{1 + 2 y}{2}

Schritte zur Lösung

\int_{y}^{y + 1} x + y 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_{y}^{1 + y} y d x + \int_{y}^{1 + y} x d x

\int_{y}^{1 + y} y d x = y

\int_{y}^{1 + y} x d x = \frac{1 + 2 y}{2}

= y + \frac{1 + 2 y}{2}

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

\int_{0}^{11} 2000 - 12 x^{2} d x

\int_{0}^{10} \frac{x}{50} d x

\int_{0}^{4} \frac{1}{1 + 2 x} d x

\int_{3}^{\infty} 2 e^{- s t} d t