integral from-1 to 2 of |x-1|sin(x)

解答

\int_{- 1}^{2} ∣ x - 1 ∣ sin (x) d x

- 3 sin (1) + 2 cos (1) - cos (2) + sin (2)

十进制

- 0.11836 \dots

求解步骤

\int_{- 1}^{2} ∣ x - 1 ∣ sin (x) d x

拿掉绝对值

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

\int_{- 1}^{1} (- x + 1) sin (x) d x = - 2 sin (1) + 2 cos (1)

\int_{1}^{2} (x - 1) sin (x) d x = - cos (2) + sin (2) - sin (1)

= - 2 sin (1) + 2 cos (1) - cos (2) + sin (2) - sin (1)

化简

= - 3 sin (1) + 2 cos (1) - cos (2) + sin (2)

\int_{0}^{2} sin (x^{2}) x d x

\int_{0}^{- 1} x^{3} 1 - x^{2} d x

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

\int_{0}^{2} 3 x + 1 d x

\int_{0}^{3} (x^{2} + 5) d x