23H.8 定积分对称性

定积分计算对称性 ¶

\(h(x)是奇函数,\displaystyle\int_{-a}^ah(x)\mathrm{d}x=0\)

\(h(x)是偶函数,\displaystyle\int_{-a}^ah(x)\mathrm{d}x=2\displaystyle\int_{0}^ah(x)\mathrm{d}x\)

\(h(x)+h(-x) = 2b(关于(0,b))中心对称\Rightarrow\displaystyle\int_{-a}^a(h(x)-b)\mathrm{d}x = 0\Rightarrow \displaystyle\int_{-a}^ah(x)\mathrm{d}x = 2ab\)

\(一般函数 h(x) = \dfrac{h(x)-h(-x)}{2}+\dfrac{h(x)+h(-x)}{2}\)

8. (2023 年微积分 (H) I 期末考试第 8 题 ) 记 \(a = \dfrac{\pi}{6}\)，计算

\(设f(x) = (\sin x)^2,g(x) = \dfrac{1}{1+e^{-x}},h(x) = f(x)g(x)\)

\(\begin{cases} f(-x) = f(x) \\ g(x)+g(-x) = 1 \end{cases}\Rightarrow h(x) + h(-x) = (\sin x)^2\)

\(\displaystyle\int_{-a}^ah(x)\mathrm{d}x = \displaystyle\int_{-a}^a\dfrac{h(x)+h(-x)}{2}\mathrm{d}x = \displaystyle\int_0^a(\sin x)^2\mathrm{d}x\)

\(= \displaystyle\int_0^a\dfrac{1-\cos 2x}{2}\mathrm{d}x = \dfrac{a}{2} - \dfrac{1}{4}\sin 2a = \dfrac{\pi}{12} - \dfrac{\sqrt{3}}{8}\)