넓이(여러 가지 함수)

All enclosed regions have the same area, with regions being the reflection of adjacent regions. One region is formed on $[\pi/4,5\pi/4]$, with area $2\sqrt{2}$.

On regions such as $[\pi/6,5\pi/6]$, the area is $3\sqrt{3}/2$. On regions such as $[-\pi/2,\pi/6]$, the area is $3\sqrt{3}/4$.

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,

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

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 $0-(y^2+y)$.

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

We need to figure out which curve is on top, when. To do this, set $h(x) = 3x - x\sqrt{25-x^2}$. If $h(x) > 0$, then $y=3x$ is the top curve; if $h(x)<0$, then $y=x\sqrt{25-x^2}$ is the top curve.

$$\begin{aligned} h(x) &= 3x - x\sqrt{25-x^2} = x\left[3-\sqrt{25-x^2}\right] \\ \text{We only care about values of $x$ in $[0,4]$, so $x$ is nonnegative. Then $h(x)$ is positive when:} \\ 3&> \sqrt{25-x^2}\\ 9&> 25-x^2\\ x^2 & > 16\\ x&> 4 \end{aligned}$$

That is, $h(x)$ is never positive over the interval $[0,4]$. So, $y = x \sqrt{25-x^2}$ lies above $y=3x$ for all $0\le x\le 4$.

The area we need to calculate is therefore:

$$\begin{aligned} A &= \int_0^4 \left[x \sqrt{25-x^2} - 3x\right]\,\mathrm{d}x \\ &= \int_0^4 x \sqrt{25-x^2}\,\mathrm{d}x - \int_0^4 3x\,\mathrm{d}x \\&= A_1 - A_2. \end{aligned}$$

To evaluate $A_1$, we use the substitution $u(x) = 25-x^2$, for which $\mathrm{d}u = u'(x)\,\mathrm{d}x= -2x\,\mathrm{d}x$; and $u(4)=25-4^2=9$ when $x=4$, while $u(0)=25-0^2=25$ when $x=0$. Therefore

$$\begin{aligned} A_1 &= \int_{x=0}^{x=4} x \sqrt{25-x^2}\,\mathrm{d}x = -\frac{1}{2} \int_{u=25}^{u=9} \sqrt{u}\,\mathrm{d}u = \left[-\frac{1}{3} u^{3/2} \right]_{25}^{9} = \frac{125 - 27}{3} = \frac{98}{3} \end{aligned}$$

For $A_2$ we use the antiderivative directly:

$$A_2 = \int_0^4 3x\,\mathrm{d}x =\left[ \frac{3x^2}{2} \right]_0^4 = 24$$

Therefore the total area is:

$$\begin{aligned} A = \frac{98}{3} - 24 = \frac{26}{3} \end{aligned}$$

Both of these quadratic functions pass through the points $(-1,0)$ and $(1,0)$, and thus these are the two intersection points for all positive values of $a$. In addition, since $1 - x^2 > ax^2 - a$ on $-1 < x < 1$, it follows that the area between the curves is given by

$$A = \int_{-1}^{1} \left( 1 - x^2 - (ax^2 - a) \right) \, dx = \int_{-1}^{1} \left( (1+a) - (1+a)x^2 \right) \, dx\,.$$ Evaluating the integral, we find that

$$\begin{aligned}A &= \int_{-1}^{1} \left( 1 + a - (1+a)x^2 \right) \, dx \\ &= \left. (1+a)x - \frac{1+a}{3}x^3 \right|_{-1}^{1} \\ &= (1+a)(1) - \frac{1+a}{3}(1)^3 - \left( (1+a)(-1) - \frac{1+a}{3}(-1)^3 \right) \\ &= 2(1+a) - \frac{2(1+a)}{3} \\ &= \frac{4(1+a)}{3}\end{aligned}\,.$$ We want to know for which value(s) of $a$ it follows that $A = 2$. Thus, we solve

$$2 = \frac{4(1+a)}{3}\,,$$ from which it follows that $a = \frac{1}{2}$. Hence, in the figure below, the area between the two curves is exactly $A = 2$.