每日题意[1256]割补

已知函数 $f(x)=\arcsin(\sin x)$,则函数 $g(x)=x\cdot |f(x)|$($x\in [0,10\pi]$)与 $x$ 轴围成的图形的面积是(       )

A.$\dfrac{665\pi^3}4$

B.$\dfrac{665\pi^3}2$

C.$\dfrac{25\pi^3}2$

D.$25\pi^3$

答案    C.

解析    根据题意,函数 $f(x)$ 是以 $2\pi$ 为周期的函数且\[f(x)=\begin{cases} x,&x\in \left[-\dfrac{\pi}2,\dfrac{\pi}2\right),\\ \pi-x,&x\in\left[\dfrac{\pi}2,\dfrac{3\pi}2\right),\end{cases}\]进而可得\[f(x)=\begin{cases} x-2k\pi,&x\in\left[2k\pi-\dfrac{\pi}2,2k\pi+\dfrac{\pi}2\right),\\ (2k+1)\pi-x,&x\in\left[2k\pi+\dfrac{\pi}2,2k\pi+\dfrac{3\pi}2\right),\end{cases}\]于是函数\[g(x)=\begin{cases} x^2-2k\pi x,&x\in\left[2k\pi,2k\pi+\dfrac{\pi}2\right),\\ (2k+1)\pi x-x^2,&x\in\left[2k\pi+\dfrac{\pi}2,2k\pi+\pi\right),\\ x^2-(2k+1)\pi x,&x\in\left[2k\pi+\pi,2k\pi+\dfrac{3\pi}2\right),\\ (2k+2)\pi x-x^2,&x\in\left[2k\pi+\dfrac{3\pi}2,2k\pi+2\pi\right),\end{cases}\]于是\[\begin{split}\int_{2k\pi}^{(2k+1)\pi} g(x){ {\rm d}} x&=\int_{2k \pi}^{\left(2k+\frac 12\right)\pi}\left[(x^2-2k\pi x)+(2k+1)\pi \left(x+\dfrac{\pi}2\right)-\left(x+\dfrac{\pi}2\right)^2{ {\rm d}} x\right]\\ &=\int_{2k \pi}^{\left(2k+\frac 12\right)\pi}\left(k+\dfrac 14\right)\pi^2{ {\rm d}} x\\ &=\left(k+\dfrac 14\right)\cdot \dfrac{\pi^3}2,\end{split}\]类似的,有\[\begin{split}\int_{(2k+1)\pi}^{(2k+2)\pi} g(x){ {\rm d}} x&=\int_{(2k+1) \pi}^{\left(2k+\frac 32\right)\pi}\left[x^2-(2k+1)\pi x+(2k+2)\pi \left(x+\dfrac{\pi}2\right)-\left(x+\dfrac{\pi}2\right)^2{ {\rm d}} x\right]\\ &=\int_{(2k+1) \pi}^{\left(2k+\frac 32\right)\pi}\left(k+\dfrac 34\right)\pi^2{ {\rm d}} x\\ &=\left(k+\dfrac 34\right)\cdot \dfrac{\pi^3}2,\end{split}\]因此所求面积为\[\sum_{k=0}^4(2k+1)\cdot \dfrac{\pi^3}2=\dfrac{25\pi^3}2.\]

备注    本质即割补法.如图,为 $g(x)$ 在 $[0,4\pi]$ 上的图象.

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