넓이(여러 가지 함수)

$9/2$

$9/4$

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

1

5

4

$21/2$

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 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

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

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 $1$.

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

If the curves $y=f(x)=x$ and $y=g(x)=3x-x^2$ intersect at $(x,y)$, then

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

Furthermore, $g(x)-f(x) = 2x-x^2 = x(2-x)$ is positive for all $0\le x\le 2$. That is, the curve $y=3x-x^2$ lies above the line $y=x$ for all $0\le x\le 2$.

We therefore evaluate the integral:

$$\begin{aligned} \int_0^2 \big[ (3x-x^2) - x \big] \,\mathrm{d}x = \int_0^2 [2x-x^2]\,\mathrm{d}x = \bigg[x^2 - \frac{x^3}{3}\bigg]^{2}_{0} = \bigg[ 4-\frac{8}{3} \bigg] -0 = \frac{4}{3} \end{aligned}$$

For our computation, we will need an antiderivative of $x^2\sqrt{x^3+1}$, which can be found using the substitution $u=x^3+1$, $\mathrm{d}u = 3x^2\,\mathrm{d}x$:

$$\begin{aligned} \int x^2\sqrt{x^3+1} \, \mathrm{d}x = \int \sqrt u \cdot \frac13\,\mathrm{d}u = \frac13\int u^{1/2}\,\mathrm{d}u = \frac13\cdot \frac{u^{3/2}}{3/2}+C = \frac29(x^3+1)^{3/2} + C. \end{aligned}$$

The two functions $f(x)$ and $g(x)$ are clearly equal at $x=0$. If $x\ne0$, then the functions are equal when

$$\begin{aligned} 3x^2 &= x^2\sqrt{x^3+1} \\ 3 &= \sqrt{x^3+1} \\ 9 &= x^3+1 \\ 8 &= x^3 \\ 2 &= x. \end{aligned}$$

The function $g(x)=3x^2$ is the larger of the two on the interval $[0,2]$, as can be seen by plugging in $x=1$, say, or by observing that when $x$ is very small $f(x)=x^2\sqrt{x^3+1}\approx x^2$ and $g(x)=3x^2$.

The area in question is therefore:

$$\begin{aligned} \int_0^2 \big( 3x^2 - x^2\sqrt{x^3+1} \big) \, \mathrm{d}x &= \bigg( {x^3} - \frac29(x^3+1)^{3/2} \bigg) \bigg|_0^2 \\ &= \bigg( 2^3 - \frac2 9(2^3+1)^{3/2} \bigg) - \bigg( 0^3 - \frac29(0^3+1)^{3/2} \bigg) \\ &= \bigg( 8 - 6 \bigg) - \bigg( 0 - \frac 2 9 \bigg) =\frac{20}9. \end{aligned}$$

First, let's figure out what our curve $x=y^2+y=y(y+1)$ looks like.

• The curve intercepts the $y$-axis when $y=0$ and $y=-1$.
• The $x$-values of the curve are negative when $-1<y<0$, and positive elsewhere.

This leads to the figure below. We're evaluating the area from $y=-1$ to $y=0$. Since $y^2+y$ is negative there, the length of our (horizontal) slices are $0-(y^2+y)$.

$$\begin{aligned} \text{Area}=\int_{-1}^0\big(0-(y^2+y)\big)\,\mathrm{d}y = -\bigg[\frac{y^3}{3}+\frac{y^2}{2}\bigg]_{-1}^0 =-\frac13+\frac12 =\frac{1}{6}\quad \smash{  } \end{aligned}$$

Let's begin by sketching our region. Note that $y=\sqrt{1-x^2}$ and $y=\sqrt{9-x^2}$ are the top halves of circles centred at the origin with radii 1 and 3, respectively.

Our region is the difference of two quarter-circles, so we find its area using geometry: $$\text{Area}=\frac{1}{4}\left(\pi\cdot 3^2\right)-\frac{1}{4}\left(\pi\cdot 1^2\right)=2\pi$$