integral from-2 to 0 of (x+1)

Lösung

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

0

Schritte zur Lösung

\int_{- 2}^{0} x + 1 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_{- 2}^{0} x d x + \int_{- 2}^{0} 1 d x

\int_{- 2}^{0} x d x = - 2

\int_{- 2}^{0} 1 d x = 2

= - 2 + 2

Vereinfache

= 0

\int_{0}^{T} x d x

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

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\int_{4}^{5} 2 x d x