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

Lösung

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

- 2

Schritte zur Lösung

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

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

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

= - 1 - 1

Vereinfache

= - 2

f^{^{''}} (x) = - f (x), f (0) = 0, f^{^{'}} (0) = 1

n = 1 \sum \infty e^{- 6 n} - e^{- 6 (n + 1)}

\int (4 x - 3) 5 d x

\int_{- 2}^{6} 4 x - x^{2} - 12 d x

d er i v a t i v e 3 x^{2} e^{x} + e^{x} (x^{3} + 7)