미적분의 기본정리

(a) The net signed area enclosed by $f$ and the $x$-axis on the interval $[2,5]$ is $\int_2^5 f(x) \, dx$. The figure below shows a graph of $f$ on the interval $[0,5]$. To calculate the definite integral of $f$ on the interval $[2,5]$, we consider the integrals on the subintervals $[2,3]$ and $[3,5]$. Doing so and using the Fundamental Theorem of Calculus, $$\begin{aligned}\int_2^5 f(x) \, dx &= \int_2^3 f(x) \, dx + \int_3^5 f(x) \, dx \\ &= \int_2^3 -x+3 \, dx + \int_3^5 x^2-8x+15 \, dx \\ &= \left. \left( -\frac{1}{2}x^2 + 3x \right)\right|_2^3 + \left. \left( \frac{1}{3}x^3 - 4x^2 + 15x \right)\right|_3^5 \\ &= \left(\frac{9}{2}-4\right) + \left( \frac{50}{3}-18 \right) \\ &= \frac{127}{6} - 22 = -\frac{5}{6}\end{aligned}\,.$$ (b) To find the average value of $f$ on $[0,5]$ we need to first calculate $$\int_0^5 f(x) \, dx\,.$$ Using our work in (a) and considering the subintervals $[0,2]$ and $[2,5]$, $$\begin{aligned}\int_0^5 f(x) \, dx &= \int_0^2 f(x) \, dx + \int_2^5 f(x) \, dx \\ &= \int_0^2 -x^2 + 2x + 1 \, dx - \frac{5}{6} \\ &= \left. \left( -\frac{1}{3}x^3 + x^2 + x \right)\right|_0^2 - \frac{5}{6} \\ &= \frac{10}{3} - \frac{5}{6} \\ &= \frac{5}{2}\end{aligned}\,.$$ Therefore, the average value of $f$ on $[0,5]$ is $$\displaystyle f_{\operatorname{AVG} [0,5]} = \frac{1}{5-0} \int_0^5 f(x) \, dx = \left(\frac{1}{5}\right) \left(\frac{5}{2}\right) = \frac{1}{2}\,.$$ (c) Since $\int_0^5 f(x) \, dx = \frac{5}{2}$, we need a function $g$ that matches $f$ but that is also defined on $[5,7]$ so that $\int_5^7 g(x) \, dx = -\frac{5}{2}$. To preserve continuity, we will choose a formula for $g$ that also satisfies $g(5) = 0$. It's natural to pick a linear function whose graph encloses $5/2$ units of area, so one that has a slope of $5/4$. To check, we compute $$\begin{aligned}\int_5^7 -\frac{5}{4}(x - 5) \, dx &= -\left(\frac{5}{4}\right) \left. \left(\frac{x^2}{2} - 5x\right) \right|_5^7 \\ &= -\left(\frac{5}{4}\right) \left( \frac{49-25}{2} - 5(7-5) \right) \\ &= -\left(\frac{5}{4}\right) \left( 12 - 10 \right) \\ &= -\left(\frac{5}{4}\right) \left( 2 \right) = -\frac{5}{2}\end{aligned}\,.$$ So $g$ defined so that $g(x) = f(x)$ for $0 \le x < 5$ and $g(x) = -\frac{5}{4}(x - 5)$ on $5 \le x \le 7$ has the desired property. A graph of $g$ is shown in the figure below.