integral from 0 to 4 of sqrt(x)+e^{-x}-1

Lösung

\int_{0}^{4} x + e^{- x} - 1 d x

\frac{7}{3} - \frac{1}{e ^{4}}

Dezimale

2.31501 \dots

Schritte zur Lösung

\int_{0}^{4} x + e^{- 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_{0}^{4} x d x + \int_{0}^{4} e^{- x} d x - \int_{0}^{4} 1 d x

\int_{0}^{4} x d x = \frac{16}{3}

\int_{0}^{4} e^{- x} d x = 1 - \frac{1}{e ^{4}}

\int_{0}^{4} 1 d x = 4

= \frac{16}{3} + 1 - \frac{1}{e ^{4}} - 4

Vereinfache

\frac{16}{3} + 1 - \frac{1}{e ^{4}} - 4 : \frac{7}{3} - \frac{1}{e ^{4}}

= \frac{7}{3} - \frac{1}{e ^{4}}

\int_{0}^{3} 3 x^{2} + x - 2 d x

\int_{0}^{1} x (x - x^{3}) d x

\int_{0}^{4} 3 e^{t} d t

\int_{- 2}^{2} 1 6^{\frac{x}{2}} d x

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