integral from 1 to 2 of xcos(x)

解答

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

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

十进制

0.02067 \dots

求解步骤

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

使用分布积分法

= [x sin (x) - \int sin (x) d x]_{1}^{2}

\int sin (x) d x = - cos (x)

= [x sin (x) - (- cos (x))]_{1}^{2}

化简

= [x sin (x) + cos (x)]_{1}^{2}

带入上下限计算后得:

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

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

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