integral from 3 to 9 of |sin(x)|

解答

\int_{3}^{9} ∣ sin (x) ∣ d x

4 + cos (3) - cos (9)

十进制

3.92113 \dots

求解步骤

\int_{3}^{9} ∣ sin (x) ∣ d x

拿掉绝对值

= \int_{3}^{π} sin (x) d x + \int_{π}^{2 π} - sin (x) d x + \int_{2 π}^{9} sin (x) d x

\int_{3}^{π} sin (x) d x = 1 + cos (3)

\int_{π}^{2 π} - sin (x) d x = 2

\int_{2 π}^{9} sin (x) d x = - cos (9) + 1

= 1 + cos (3) + 2 - cos (9) + 1

化简

= 4 + cos (3) - cos (9)

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

\int_{0}^{1} \frac{x ^{2} + 1}{x ^{4} + 1} d x

\int_{- 1}^{1} (∣ x ∣ + x) e^{- ∣ x ∣} d x

\int_{2}^{5} 3 x^{2} d x

\int_{π}^{2 π} (x - 2 sin (x)) d x