integral from 1 to 2.8 of ln(x^2)

解答

\int_{1}^{2.8} ln (x^{2}) d x

2.16586 \dots

求解步骤

\int_{1}^{2.8} ln (x^{2}) d x

使用分布积分法

= [x ln (x^{2}) - \int 2 d x]_{1}^{2.8}

\int 2 d x = 2 x

= [x ln (x^{2}) - 2 x]_{1}^{2.8}

使用对数运算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

假定

x \geq 0

= [2 x ln (x) - 2 x]_{1}^{2.8}

带入上下限计算后得:

2.16586 \dots

= 2.16586 \dots

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