integral from-4 to x of (t-1)

Lösung

\int_{- 4}^{x} (t - 1) d t

\frac{x ^{2}}{2} - x - 12

Schritte zur Lösung

\int_{- 4}^{x} t - 1 d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 4}^{x} t d t - \int_{- 4}^{x} 1 d t

\int_{- 4}^{x} t d t = \frac{x ^{2}}{2} - 8

\int_{- 4}^{x} 1 d t = x + 4

= \frac{x ^{2}}{2} - 8 - (x + 4)

Vereinfache

= \frac{x ^{2}}{2} - x - 12

\int_{0}^{l n (4)} e^{5 x} d x

\int_{1}^{4} 7 u d u

\int_{0}^{4} e^{- \frac{x}{2}} d x

\int_{- 2}^{3} (2 - ∣ x ∣) d x

\int_{- 1}^{1} x^{3} - 4 x d x