넓이(여러 가지 함수)

We will compute the area by using thin vertical strips, as in the sketch below:

By looking at the sketch above, we guess the line $y = 4 + 2\pi - 2x$ intersects the curve $y = 4 + \pi \sin x$ when $x=\frac{\pi}{2},$ $x=\pi$, and $x=\frac{3\pi}{2}$. Let's make sure these are correct by plugging them into the two equations, and making sure the $y$-values match:

$$\begin{array}{|c|c|c|c|}\hline x & 4+2\pi-2x & 4+\pi\sin(x) & \text{match?} \\ \hline \frac{\pi}{2} & 4+\pi & 4+\pi & \text{\checkmark} \\ \hline \pi & 4 & 4 & \text{\checkmark} \\ \hline \frac{3\pi}{2} & 4-\pi & 4-\pi & \text{\checkmark} \\ \hline \end{array}$$

Also from the sketch, we see that:

• When $\frac{\pi}{2} \le x \le \pi$, the top of the strip is at $y = 4 + \pi \sin x$ and the bottom of the strip is at $y = 4 + 2\pi - 2x$. So the strip has height $\big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]$ and width $\mathrm{d}x$, and hence area $\big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]\mathrm{d}x$.
• When $\pi \le x \le \frac{3\pi}{2}$, the top of the strip is at $y = 4 + 2\pi - 2x$ and the bottom of the strip is at $y = 4 + \pi \sin x$. So the strip has height $\big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]$ and width $\mathrm{d}x$, and hence area $\big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]\mathrm{d}x$.

Now we can calculate:

$$\begin{aligned} \text{Area} &= \int_{\pi/2}^\pi \big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]\ \mathrm{d}x +\int^{3\pi/2}_\pi \big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]\ \mathrm{d}x\\ &= \int_{\pi/2}^\pi \big[\pi \sin x- 2\pi + 2x\big]\ \mathrm{d}x +\int^{3\pi/2}_\pi \big[2\pi - 2x- \pi \sin x\big]\ \mathrm{d}x\\ &=\Big[-\pi \cos x- 2\pi x + x^2\Big]_{\pi/2}^\pi +\Big[2\pi x - x^2+ \pi \cos x\Big]^{3\pi/2}_\pi\\ &=\left[\pi-\pi^2+\frac{3}{4}\pi^2\right] +\left[\pi^2-\frac{5}{4}\pi^2+\pi\right]\\ &=2\Big[\pi-\frac{1}{4}\pi^2\Big] \end{aligned}$$