integral from 1 to 2 of 2sin(x)+cos(x)

解答

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

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

十进制

1.98072 \dots

求解步骤

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

使用积分加法定则:

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

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

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

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

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

\int_{2}^{3} x^{\frac{2}{3}} (- 1 + x) d x

\int_{0}^{4} (x - 1) d x

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

\int_{- 2}^{3} (3 x^{4} - 5 x^{3} - x - 6) d x

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