넓이(여러 가지 함수)

First, here is a sketch of the region. We are not asked for it, but it is crucial for understanding the question.

The two curves $y=x+2$ and $y=x^2$ cross at $(2,4)$. The area of the part between them with $0\le x\le 2$ is:

$$\begin{aligned} \int_0^2 \big[x+2-x^2\big]\,\mathrm{d}x=\Big[\frac{1}{2} x^2+2x-\frac{1}{3}x^3\Big]_0^2 =2+4-\frac{8}{3}=\frac{10}{3} \end{aligned}$$

The area of the part between the two curves with $2\le x\le 3$ is:

$$\begin{aligned} \int_2^3 \big[x^2-(x+2)\big]\,\mathrm{d}x=\Big[\frac{1}{3}x^3-\frac{1}{2} x^2-2x\Big]_2^3 =9-\frac{9}{2}-6-\frac{8}{3}+2+4=\frac{11}{6} \end{aligned}$$

The total area is $\dfrac{10}{3}+\dfrac{11}{6}=\dfrac{31}{6}$.

Let's begin by sketching our region. Note that $y=\sqrt{9-x^2}$ is the top half of a circle centred at the origin with radius 3, while $y=\sqrt{1-(x-1)^2}$ is the top half of a circle of radius 1 centred at $(1,0)$.

Note $y=x$ intersects $y=\sqrt{1-(x-1)^2}$ at $(1,1)$, the highest part of the smaller half-circle.

We can easily take the area of triangles and sectors of circles. With that in mind, we cut up our region the following way:

• The desired area is $A_3-(A_1+A_2)$.
• $A_1$ is the area of right a triangle with base 1 and height 1, so $A_1 = \frac{1}{2}$.
• $A_2$ is the area of a quarter circle of radius 1, so $A_2=\frac{\pi}{4}$.
• $A_3$ is the area of an eighth of a circle of radius 3, so $A_3 = \frac{9\pi}{8}$

So, the area of our region is $\dfrac{9\pi}{8} - \dfrac{1}{2}-\dfrac{\pi}{4}=\dfrac{7\pi}{8}-\dfrac{1}{2}$.

The first function is a cubic, with intercepts at $x=0,\pm2$. The second is a straight line with a positive slope.

We need to figure out what these functions look like in relation to one another, so let's find their points of intersection.

$$\begin{aligned} x(x^2-4)&=x-2\\ x(x+2)(x-2)&=x-2\\ \boxed{x-2=0} \quad\text{or}\quad x(x+2)&=1\\ x^2+2x-1&=0\\ x &= \dfrac{-2\pm\sqrt{4-4(1)(-1)}}{2}\\ x&=\boxed{-1\pm \sqrt{2}} \end{aligned}$$

So, our three points of intersection are when $x=2$ and when ${x=-1\pm\sqrt{2}}$. We note $$-1-\sqrt{2} < -1+\sqrt{2} < -1+\sqrt{4}< 2\ .$$ So, we need to see which function is on top over the two intervals $\left[-1-\sqrt{2},-1+\sqrt{2}\right]$ and $\left[-1+\sqrt{2},2\right]$. It suffices to check points in these intervals.

$$\begin{array}{|c|c|c|c|}\hline x & x(x^2-4) & x-2 & \text{top function:} \\ \hline \text{0} & \text{0} & -2 & x(x^2-4) \\ \hline \text{1} & \text{-3} & -1 & x-2 \\ \hline \end{array}$$

Since 0 is in the interval $\left[-1-\sqrt{2},-1+\sqrt{2}\right]$, $x(x^2-4)$ is the top function in that interval. Since 1 is in the interval $\left[-1+\sqrt{2},2\right]$, $x-2$ is the top function in that interval. Now we can set up the integral to evaluate the area:

$$\begin{aligned} \text{Area}&=\int_{-1-\sqrt{2}}^{-1+\sqrt{2}}\left[x(x^2-4) - (x-2)\right]\ \mathrm{d}x \quad&&+ \quad &&\int_{-1+\sqrt{2}}^{2}\left[(x-2)-x(x^2-4)\right]\ \mathrm{d}x\\ &=\int_{-1-\sqrt{2}}^{-1+\sqrt{2}}\left[x^3-5x+2\right]\ \mathrm{d}x \quad&&+ \quad &&\int_{-1+\sqrt{2}}^{2}\left[-x^3+5x-2\right]\ \mathrm{d}x \\&=\left[\frac{1}{4}x^4 - \frac{5}{2}x^2+2x\right]_{-1-\sqrt{2}}^{-1+\sqrt{2}} \quad&&+ \quad &&\left[-\frac{1}{4}x^4 + \frac{5}{2}x^2-2x\right]_{-1+\sqrt{2}}^{2} \\ \text{After some taxing but rudimentary algebra:} \\ &=\left(8\sqrt{2}\right)+\left(4\sqrt{2}-\frac{13}{4}\right)=12\sqrt{2}-\frac{13}{4} \end{aligned}$$