넓이(여러 가지 함수)

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:

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

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 $\dfrac{10}{3}+\dfrac{11}{6}=\dfrac{31}{6}$.

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:

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

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:

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

(a) The point of intersection of the graphs occurs when $y(y - 2) = -(y - 1)(y - 3)$, and thus we need to find where $y^2 - 2y = -(y^2 - 4y + 3)$ or $2y^2 - 6y + 3 = 0$. By the quadratic formula, it follows that $$y = \frac{6 \pm \sqrt{(-6)^2 - (4)(2)(3)}}{(2)(2)} = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{3 \pm \sqrt{3}}{2}\,.$$ Since the curve $x = -(y - 1)(y - 3)$ lies to the right (has a greater $x$ value) of $x = y(y - 2)$ between $y = \frac{3 - \sqrt{3}}{2} \approx 0.63$ and $y = \frac{3 + \sqrt{3}}{2} \approx 2.37$ as shown in the figure below, we see that the area between the two curves is given by the integral $$\int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -(y - 1)(y - 3) - y(y - 2) \ dy\,.$$ Evaluating this integral, we find that $$\begin{aligned}A &= \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -2y^2 + 6y - 3 \ dy \\ &= \left. \left(-\frac{2}{3}y^3 + 3y^2 - 3y \right) \right|_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} \\ &= \left[-\frac{2}{3} \left(\frac{3 + \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 + \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 + \sqrt{3}}{2}\right)\right] \\ &\qquad - \left[-\frac{2}{3} \left(\frac{3 - \sqrt{3}}{2}\right)^3 + 3\left(\frac{3 - \sqrt{3}}{2}\right)^2 - 3\left(\frac{3 - \sqrt{3}}{2}\right)\right] \\ &= -\frac{1}{12}(54+30\sqrt{3}) + \frac{1}{12}(54-30\sqrt{3}) + \frac{3}{4}(12+6\sqrt{3}) \\ &\qquad - \frac{3}{4}(12-6\sqrt{3}) - \frac{3}{2}(3+\sqrt{3}) + \frac{3}{2}(3-\sqrt{3}) \\ &= -5\sqrt{3} + 9\sqrt{3} - 3\sqrt{3} \\ &= \sqrt{3}\end{aligned}\,.$$ (b) As the figure below shows, the graph of $y = \sin(x)$ lies above the graph of $y = \cos(x)$ on this interval. So the area between the two curves is given by $$\begin{aligned}A &= \int_{\pi/4}^{3\pi/4} \sin(x) - \cos(x) \ dx \\ &= \left. \left(-\cos(x) - \sin(x) \right) \right|_{\pi/4}^{3\pi/4} \\ &= \left[-\cos\left(\frac{3\pi}{4}\right) - \sin\left(\frac{3\pi}{4}\right) \right] - \left[-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right) \right] \\ &= \left[ \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \right] + \left[ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right] \\ &= \sqrt{2}\end{aligned}\,.$$ (c) We can rewrite $y = 2x - 1$ as $x = \frac{y+1}{2}$. The two curves intersect when $y^2 - y - 2 = \frac{y+1}{2}$. It follows that $2y^2 - 2y - 4 = y+1$, so $2y^2 - 3y - 5 = 0$ and thus factoring, $(2y-5)(y+1) = 0$. Hence the two curves intersect when $y = -1$ or $y = \frac{5}{2}$. The figure below shows that $x = \frac{y+1}{2}$ lies to the right of $x = y^2 - y - 2$on this interval, so the area of the region between the two curves is $$\begin{aligned}A &= \int_{-1}{5/2} \frac{y+1}{2} - (y^2-y-2) \ dy \\ &= \int_{-1}{5/2} -y^2 + \frac{3}{2}y + \frac{5}{2} \ dy \\ &= \left. \left( -\frac{y^3}{3} + \frac{3}{4}y^2 + \frac{5}{2}y \right) \right|_{-1}{5/2} \\ &= -\frac{1}{3}\left[\left(\frac{5}{2}\right)^3 + 1 \right] + \frac{3}{4}\left[\left(\frac{5}{2}\right)^2 - 1 \right] + \frac{5}{2}\left[\left(\frac{5}{2}\right) + 1 \right] \\ &= -\frac{133}{24} + \frac{63}{16} + \frac{35}{4} \\ &= \frac{343}{48}\end{aligned}\,.$$ (d) A representative picture (with $m = 2$) is shown in the figure below. The two curves intersect when $x^2 - 1 = mx$, or when $x^2 - mx - 1 = 0$. By the quadratic formula, it folows that $$x = \frac{m \pm \sqrt{m^2-4(1)(-1)}}{2} = \frac{m \pm \sqrt{m^2+4}}{2}\,.$$ If we integrate with vertical slices, then the graph of $y = mx$ is always above the graph of $y = x^2 - 1$ and so the area of the region between the two graphs is $$\begin{aligned}A &= \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx \\ &= \left. \left(-\frac{x^3}{3} + m\frac{x^2}{2} + x\right) \right|_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} \\ &= - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2 \\ &\qquad - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right)\end{aligned}\,.$$