넓이(여러 가지 함수)

(a) The average value of $f$ on the interval $[0,\sqrt{2}]$ is $$\begin{aligned}r &= \frac{1}{\sqrt{2}} \int_0^{\sqrt{2}} 2-x^2 \, dx \\ &= \left. \frac{1}{\sqrt{2}}\left(2x - \frac{x^3}{3}\right) \right|_0^{\sqrt{2}} \\ &= \frac{1}{\sqrt{2}}\left( 2\sqrt{2} - \frac{\sqrt{2}^3}{3} \right) \\ &= 2 - \frac{2}{3} \\ &= \frac{4}{3}\end{aligned}\,.$$ (b) The intersection point of $y=f(x)$ and $y=r$ occurs, for positive $x$, when $2-x^2 = \frac{4}{3}$, from which it follows that $x^2 = \frac{2}{3}$ and thus $x = \frac{\sqrt{6}}{3}$. A graph of $f$, $y = r$, and the intersection point is shown in the following figure. (c) Let $A_1$ be the area that lies below $y = f(x)$ and above $y = r$ on the interval $\left[0, \frac{\sqrt{6}}{3}\right]$ and $A_2$ the are of the region that lies below $y = r$ and above $y = f(x)$ on $\left[\frac{\sqrt{6}}{3}, \sqrt{2}\right]$ as illustrated in the figure below. Then $$\begin{aligned}A_1 &= \int_0^{\sqrt{6}/3} (2-x^2) - \frac{4}{3} \ dx \\ &= \int_0^{\sqrt{6}/3} \frac{2}{3} - x^2 \ dx \\ &= \left. \left( \frac{2}{3}x - \frac{x^3}{3} \right) \right|_0^{\sqrt{6}/3} \\ &= \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) - \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 \\ &= \frac{2\sqrt{6}}{9} - \frac{6 \sqrt{6}}{81} \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\,.$$ and $$\begin{aligned}A_2 &= \int_{\sqrt{6}/3}^{\sqrt{2}} \frac{4}{3} - (2-x^2) \ dx \\ &= \int_{\sqrt{6}/3}^{\sqrt{2}} x^2 - \frac{2}{3} \ dx \\ &= \left. \left(\frac{x^3}{3} - \frac{2}{3}x \right) \right|_{\sqrt{6}/3}^{\sqrt{2}} \\ &= \left(\frac{2\sqrt{2}}{3} - \frac{2}{3} \sqrt{2} \right) - \left[ \left(\frac{1}{3}\right) \left(\frac{\sqrt{6}}{3} \right)^3 - \left(\frac{2}{3}\right) \left(\frac{\sqrt{6}}{3} \right) \right] \\ &= \frac{4 \sqrt{6}}{27}\end{aligned}\,.$$ Hence $A_1 = A_2$ as desired. (d) The answer is yes. Let $AV$ is the average value of a function $f$ on an interval $[a,b]$. By the definition of average value, $$AV(b-a) = \int_a^b f(x) \ dx\,,$$ which can also be viewed as the area of the rectangle with base $(b-a)$ and height $AV$. Now suppose that the area of the region above the graph of $f$ and below the line $y=AV$ on the interval $[a,b]$ is $R_1$ and the area of the region below the graph of $f$ and above the line $y=AV$ on the interval $[a,b]$ is $R_2$. The figure below illustrates that the value of the definite integral is the area of the rectangle plus the area above the rectangle minus the area below the rectangle, or $$\int_a^b f(x) \, dx = AV(b-a) + R_2 - R_1\,.$$ Having already established that $AV(b-a) = \int_a^b f(x) \ dx$, it follows that $R_2 - R_1 = 0$. So the area of the region that lies below $y = f(x)$ and above $y = AV$ is equal to the amount of area that lies below $y = AV$ and above $y = f(x)$ on $[a,b]$.