넓이(여러 가지 함수)

$\pi$

$1/2$

$2\sqrt{2}$

$1/\ln 4$

$4/3$

$1$

$5/3$

$9/2$

$9/4$

$1/12(9-2\sqrt{2})\approx 0.514$

219,000 ft$^2$

623,333 ft$^2$

The intervals of our rectangles are $[0,\frac{\pi}{4}]$, $[\frac{\pi}{4},\frac{\pi}{2}]$, $[\frac{\pi}{2},\frac{3\pi}{4}]$, and $[\frac{3\pi}{4},\pi]$. Since we're taking a left Riemann sum, we find the height of the rectangles at the left endpoints of the intervals.

• $x = 0$: The distance from $\cos 0$ to $\sin 0$ is 1, so our first rectangle has height 1.
• $x = \frac{\pi}{4}$: The distance from $\cos \frac{\pi}{4}$ to $\sin \frac{\pi}{4}$ is 0, so our second rectangle has height 0.
• $x = \frac{\pi}{2}$: The distance from $\cos \frac{\pi}{2}$ to $\sin \frac{\pi}{2}$ is 1, so our third rectangle has height 1.
• $x = \frac{3\pi}{4}$: The distance from $\cos \frac{3\pi}{4}$ to $\sin \frac{3\pi}{4}$ is $\sin(3\pi/4)-\cos(3\pi/4) =\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)=\sqrt{2}$, so our fourth rectangle has height $\sqrt{2}$.

So, our approximation for the area between the two curves is $$\frac{\pi}{4}\left(1+0+1+\sqrt{2}\right)=\frac{\pi}{4}\left(2+\sqrt{2}\right)$$

(a) We are finding the area in the interval from $x=0$ to $x=\frac{\pi}{2}$. Since we're taking $n=5$ rectangles, our rectangles cover the following intervals: $$\left[0,\frac{\pi}{10}\right],\quad \left[\frac{\pi}{10},\frac{\pi}{5}\right],\quad \left[\frac{\pi}{5},\frac{3\pi}{10}\right],\quad \left[\frac{3\pi}{10},\frac{2\pi}{5}\right],\quad \left[\frac{2\pi}{5},\frac{\pi}{2}\right].$$

(b) We are finding the area in the interval from $y=0$ to $y=\frac{\pi}{2}$. (In general, when we switch from horizontal rectangles to vertical, the limits of integration will change–it's only coincidence that they are the same in this example.) Since we're taking $n=5$ rectangles, these rectangles cover the following intervals of the $y$-axis: $$\left[0,\frac{\pi}{10}\right],\quad \left[\frac{\pi}{10},\frac{\pi}{5}\right],\quad \left[\frac{\pi}{5},\frac{3\pi}{10}\right],\quad \left[\frac{3\pi}{10},\frac{2\pi}{5}\right],\quad \left[\frac{2\pi}{5},\frac{\pi}{2}\right].$$

The question doesn't specify which endpoints we're using. Let's use upper endpoints, to match part (a).

The area between the curve $y= \frac{1}{(2x-4)^2}$ and the $x$-axis, with $x$ running from $a=0$ to $b=1$, is exactly the definite integral of $\frac{1}{(2x-4)^2}$ with limits $0$ and $1$.

$$\begin{aligned} \text{Area}&=\int_0^1 \frac{\mathrm{d}x}{(2x-4)^2}&u=2x-4,\quad\mathrm{d}u=2\ \mathrm{d}x\\ &=\frac{1}{2}\int_{-4}^{-2}\frac{1}{u^2}\mathrm{d}u = \frac{1}{2}\left[\frac{-1}{u}\right]_{u=-4}^{u=-2}\\ &=\frac{1}{2}\Big[\frac{1}{2}-\frac{1}{4}\Big] =\frac{1}{8} \end{aligned}$$

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:

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