integral from x to 1 of 2θsin(θ)

解答

\int_{x}^{1} 2 θ sin (θ) d θ

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

求解步骤

\int_{x}^{1} 2 θ sin (θ) d θ

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{x}^{1} θ sin (θ) d θ

使用分布积分法

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

\int - cos (θ) d θ = - sin (θ)

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

化简

= 2 [- θ cos (θ) + sin (θ)]_{x}^{1}

带入上下限计算后得:

- cos (1) + sin (1) + x cos (x) - sin (x)

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

\int_{0}^{2} \frac{6}{7} (x^{2} + \frac{x y}{2}) d y

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

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