넓이(여러 가지 함수)

(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 $$\begin{aligned}A &= \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx \\ &= \left. \left(-\frac{x^3}{3} + m\frac{x^2}{2} + x\right) \right|_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} \\ &= - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2 \\ &\qquad - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right)\end{aligned}\,.$$