integral from-1 to 1 of |e^x-1|

解答

\int_{- 1}^{1} ∣ e^{x} - 1 ∣ d x

\frac{1}{e} + e - 2

十进制

1.08616 \dots

求解步骤

\int_{- 1}^{1} ∣ e^{x} - 1 ∣ d x

拿掉绝对值

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

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

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

= \frac{1}{e} + e - 2

\int_{2}^{4} 3 x^{2} d x

\int_{- 1}^{1} x + 2 x^{2} - x^{3} + 5 x^{4} d x

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

\int_{0}^{c o s (x)} e^{- t^{3}} d t

\int_{- π}^{0} 2 x^{4} d x