넓이(여러 가지 함수)

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 curves intersect when $y=x$ and $y=x^3-x$. To find these points, we set:

$$\begin{aligned} x&=x^3-x\\ 0&=x^3-2x\\ 0&=x(x^2-2)\\ 0&= x \quad\text{or}\quad 0=x^2-2 \end{aligned}$$

For $x\ge 0$, the curves intersect at $(0,0)$ and $(\sqrt{2},\sqrt{2})$.

A handy observation is that, since both curves are continuous and they do not meet each other between $x=0$ and $x=\sqrt{2}$, we don't have to worry about dividing our area into two regions: one of the functions is always on the top, and the other is always on the bottom.

Using vertical strips:

The top and bottom boundaries of the specified region are $y=T(x)=x$ and $y=B(x)=x^3-x$, respectively. So,

$$\begin{aligned} {\rm Area} = \int_0^{\sqrt{2}}\big[T(x)-B(x)\big]\ \mathrm{d}x = \int_0^{\sqrt{2}}\big[x-(x^3-x)\big]\ \mathrm{d}x = \int_0^{\sqrt{2}} 2x-x^3 \ \mathrm{d}x \end{aligned}$$

We need to find where the curves intersect.

$$\begin{aligned} \frac{x^2}{4}=y^2&=6-\dfrac{5x}{4}\\ \frac{1}{4}x^2+\frac{5}{4}x-6&=0\\ x^2+5x-24&=0\\ (x+8)(x-3)&=0\\ x=-8,\quad x&=3 \end{aligned}$$

The curves intersect at $(-8,4)$ and $(3,-\frac{3}{2})$. Using horizontal strips:

we have

$$\begin{aligned} \text{Area} = \int_{-3/2}^{4}\Big[\frac{4}{5}(6-y^2)+2y\Big]\ \mathrm{d}y \end{aligned}$$

If the curves intersect at $(x,y)$, then

$$\begin{aligned} \left(x^2\right)^2&=\left(4a\right)^2y^2 = (4a)^24ax\\ x^4&=(4a)^3 x\\ x^4&-(4a)^3x=0\\ x(&x^3-(4a)^3)=0\\ x&= 0 \quad\text{or}\quad x^3=(4a)^3 \end{aligned}$$

The curves intersect at $(0,0)$ and $(4a,4a)$. (It is also possible to find these points by inspection.) Using vertical strips:

We want the $y$-values of the functions. We write the top function as $y =\sqrt{4ax}$ (we care about the positive square root, not the negative one) and we write the bottom function as $y=\frac{x^2}{4a}$. Then we have

$$\begin{aligned} \text{Area} = \int_0^{4a}\left[\sqrt{4ax}-\frac{x^2}{4a}\right]\ \mathrm{d}x \end{aligned}$$

The curves intersect when $x=4y^2$ and $0=4y^2+12y+5 =(2y+5)(2y+1)$. So, the curves intersect at $(1,-\tfrac{1}{2})$ and $(25,-\frac{5}{2})$. Using vertical strips:

we have

$$\begin{aligned} \text{Area} = \int_1^{25}\left[-\frac{1}{12}(x+5)+\frac{1}{2}\sqrt{x}\right]\ \mathrm{d}x \end{aligned}$$