Find the average value of $g(x) = \left| x \right| – 1$ on $\left[ – 1,3 \right]$.

$\begin{array}{cl} \text{AVG.} & =\frac{{\int_{ – 1}^3 {(\left| x \right| – 1)} dx}}{{3 – ( – 1)}} = \frac{{\int_{ – 1}^0 { – x – 1} dx + \int_0^3 {(x – 1)} dx}}{4} \\ & =\frac{{(\left. { – \frac{{x^2 }}{2} – x} \right|_{ – 1}^0 ) +(\left. {\frac{{x^2 }}{2} – x} \right|_0^3 )}}{4} \\ & =\frac{{0 – ( – \frac{1}{2} + 1) + (\frac{9}{2} – 3) – 0}}{4} = \frac{1}{4} ∎ \end{array}$

Find $\int {x\sec ^2 x} dx$.

$\begin{array}{cl} \int {x\sec ^2 x} dx & =x\tan x – \int {\tan x} dx \\ & =x\tan x + \ln \left| {\cos x} \right| + C \\ & =x\tan x – \ln \left| {\sec x} \right| + C ∎ \end{array}$

Evaluate $\int_0^2 {\int_{\frac{y}{2}}^1 {ye^{x^3 } } } dxdy$.

$\begin{array}{cl} \int_0^2 {\int_{\frac{y}{2}}^1 {ye^{x^3 } } } dxdy & =\int_0^1 {e^{x^3 } } \int_0^{2x} y dydx \\ & =\int_0^1 {e^{x^3 } } \cdot \left. {\frac{{y^2 }}{2}} \right|_{y = 0}^{y = 2x} dx \\ & =\int_0^1 {e^{x^3 } } \cdot \frac{{4x^2 }}{2}dx \\&=2\int_0^1 {e^{x^3 } } x^2 dx\\&(\text{Let }u = x^3 \Rightarrow du = 3x^2 dx)\\&=\frac{2}{3}\int_0^1 {e^u } du\\&=\frac{2}{3} \cdot \left. {e^u } \right|_{u = 0}^{u = 1} = \frac{2}{3}(e – 1) ∎ \end{array}$

Find the value of $a$ and $b$ that makes the function $f(x) = \left\{ {\begin{array}{c}{\frac{{2\sin ^2 x}}{x},} & {{\text{if }}x > 0} \\ {ax + b\cos x,} & {{\text{if }}x \le 0} \\ \end{array}} \right.$ differentiable at $x=0$.

$\because\ f(x)$ is differential at $x=0$
$\therefore\ f(x)$ is continued at $x=0$
$\Rightarrow \mathop {\lim }\limits_{x \to 0^ – } f(x) = \mathop {\lim }\limits_{x \to 0^ + } f(x)$
$\Rightarrow \left\{ \begin{array}{l} \mathop {\lim }\limits_{x \to 0^ – } f(x) = \mathop {\lim }\limits_{x \to 0^ – } a\underbrace x_0 + b\underbrace {\cos x}_1 = b \\ {\mathop {\lim }\limits_{x \to 0^ + } f(x) = \mathop {\lim }\limits_{x \to 0^ + } \frac{{2\sin ^2 x}}{x} = \mathop {\lim }\limits_{x \to 0^ + } 2\underbrace {\sin x}_0 \cdot \underbrace {\frac{{\sin x}}{x}}_1 = 0} \end{array} \right.\Rightarrow b = 0$
$\Rightarrow f(x) = \left\{ {\begin{array}{c}\frac{{2\sin ^2 x}}{x}, & \text{if }x > 0 \\ {ax,} & \text{if }x \le 0 \\ \end{array}} \right.$
$\Rightarrow f'(x) = \left\{ {\begin{array}{c}\frac{{(4\sin x\cos x) \cdot x – 2\sin ^2 x}}{{x^2 }}, & \text{if }x > 0 \\ {a,} & \text{if }x \le 0 \\ \end{array}} \right.$
$\because\ f(x)$ is differentiable at $x=0$
$\therefore\ \mathop {\lim }\limits_{x \to 0^ – } f'(x) = \mathop {\lim }\limits_{x \to 0^ + } f'(x)$
$\Rightarrow \mathop {\lim }\limits_{x \to 0^ – } a = \mathop {\lim }\limits_{x \to 0^ + } (4 \cdot \underbrace {\frac{{\sin x}}{x}}_1 \cdot \underbrace {\cos x}_1 – 2 \cdot \underbrace {\frac{{\sin ^2 x}}{{x^2 }}}_1) = 4 – 2 = 2 \Rightarrow a = 2$
Thus $a=2,\ b=0 ∎$

Find the tangent line to the curve $x^2 \cos ^2 y – \sin y = 0$ at $(0,\pi)$.

$\Rightarrow 2x \cdot \cos ^2 y + x^2 \cdot 2\cos y \cdot ( – \sin y) \cdot y’ – (\cos y) \cdot y’ = 0$
At $(x,y) = (0,\pi ) \Rightarrow \left. {y’} \right|_{(0,\pi )} = 0$

the tangent line: $y=\pi ∎$

Find the volume of the solid obtained by revolving the region bounded by the curves $y = – x^2 + 4x$ and $y=x^2$ about the $x$-axis.

$\begin{array}{cl} V & =\int_0^2 {\underbrace {\pi ( – x^2 + 4x)^2 – \pi (x^2 )^2 }_{\text{area of disk}}}\underbrace {dx}_{\text{thickness}} \\ & =\pi \int_0^2 {x^4 – 8x^3 + 16x^2 – x^4 } dx \\ & =\pi \left[ { – 2x^4 + \frac{{16}}{3}x^3 } \right]_{x = 0}^{x = 2} \\&=\pi ( – 2 \times 16 + \frac{{16}}{3} \times 8)\\&=\pi (128 – 96) = \frac{{32}}{3}\pi ∎ \end{array}$

Evaluate $\mathop {\lim }\limits_{x \to 0^ + } (\sin x)^x$.

$\begin{array}{cl} \mathop {\lim }\limits_{x \to 0^ + } (\sin x)^x & =\mathop {\lim }\limits_{x \to 0^ + } e^{\ln (\sin x)^x } \\ & =e^{\mathop {\lim }\limits_{x \to 0^ + } x\ln (\sin x)} \\ & =e^{\mathop {\lim }\limits_{x \to 0^ + } \frac{{\ln (\sin x)}}{{x^{ – 1} }}} \\&\mathop = \limits^{\rm{L}} e^{\mathop {\lim }\limits_{x \to 0^ + } \frac{{\frac{{\cos x}}{{\sin x}}}}{{( – x^{ – 2} )}}}\\&=e^{\mathop {\lim }\limits_{x \to 0^ + } – \frac{{x^2 \cos x}}{{\sin x}}}\\&=e^{\mathop {\lim }\limits_{x \to 0^ + } – \underbrace {(\frac{x}{{\sin x}})}_1 \cdot \underbrace x_0 \cdot \underbrace {\cos x}_1}\\&=e^0=1 ∎ \end{array}$

Find the sum of the series $\sum\limits_{n = 3}^\infty {\frac{{\ln (1 + \frac{1}{n})}}{{(\ln n)\ln (n + 1)}}}$.

$\because\ \ln (1 + \frac{1}{n}) = \ln (\frac{{n + 1}}{n}) = \ln (n + 1) – \ln n$
$\begin{array}{cl}\therefore \sum\limits_{n = 3}^\infty {\frac{{\ln (1 + \frac{1}{n})}}{{(\ln n)\ln (n + 1)}}} & =\sum\limits_{n = 3}^\infty {\frac{{\ln (n + 1) – \ln n}}{{(\ln n)\ln (n + 1)}}} \\ & =\mathop {\lim }\limits_{k \to \infty } \sum\limits_{n = 3}^k {(\frac{1}{{\ln n}} – \frac{1}{{\ln (n + 1)}})} \\ & =\mathop {\lim }\limits_{k \to \infty } \left[ {(\frac{1}{{\ln 3}} – \frac{1}{{\ln 4}}) + (\frac{1}{{\ln 4}} – \frac{1}{{\ln 5}}) + \cdots + (\frac{1}{{\ln k}} – \frac{1}{{\ln (k + 1)}})} \right] \\&=\mathop {\lim }\limits_{k \to \infty } \left[ {\frac{1}{{\ln 3}} – \frac{1}{{\ln (k + 1)}}} \right] = \frac{1}{{\ln 3}} ∎ \end{array}$

An open rectangular box is to be constructed from material that costs \$3 /$ft^2$for the bottom and \$ 1 / $ft^2$ for its sides. Find the dimensions of the box of greatest volume that can be constructed for $36. 提示 依題意將條件列完後就和一般極值問題一樣，在微積分的考試中，有限制的極值問題先想到拉格朗日乘數法。 解答 令盒子的長$=x$, 寬$=y$, 高$=z$cost:$g(x,y,z) = 3xy + 2xz + 2yz = 36$volume:$xyz=f(x,y,z)$Let$\nabla f+\lambda \nabla g=\overset{\rightharpoonup}{0}\Rightarrow (yz,xz,xy) + \lambda (3y + 2z,3x + 2z,2x + 2y) =\overset{\rightharpoonup}{0}\Rightarrow \left\{ {\begin{array}{c} yz + 3\lambda y + 2\lambda z = 0 \cdots (1) \\ xz + 3\lambda x + 2\lambda z = 0 \cdots (2) \\ xy + 2\lambda x + 2\lambda y = 0 \cdots (3) \end{array}} \right.3xy+2xz+2yz=36 \cdots (4)\begin{array}{cl} (1)-(2) & \Rightarrow z(y – x) + 3\lambda (y – x) = 0 \\ & \Rightarrow (y – x) \cdot (z + 3\lambda ) = 0 \end{array}$若$z=-3\lambda\mathop \Rightarrow \limits^{(1)} – 3\lambda y + 3\lambda y – 6\lambda ^2 = 0 \Rightarrow \lambda = 0 \Rightarrow z = 0\mathop \Rightarrow \limits^{(3)(4)} \left\{ {\begin{array}{c}{xy = 0} & \\ {3xy = 36} & { \Rightarrow xy = 12} \end{array}} \right.\ (\to \leftarrow)\Rightarrow z \ne – 3\lambday=x\mathop \Rightarrow \limits^{(1)(3)(4)} \left\{ {\begin{array}{l} xz + 3\lambda x + 2\lambda z = 0 \cdots (6) \\ x^2 + 4\lambda x = 0 \Rightarrow x(x + 4\lambda ) = 0 \\ 3x^2 + 4xz = 36 \cdots (5) \end{array}} \right.\Rightarrow x=0\text{ or }x=-4\lambda\mathop \Rightarrow \limits^{(5)} 0 = 36\ (\to \leftarrow)\ \Rightarrow x=-4\lambda\mathop \Rightarrow \limits^{(5)(6)} \left\{ {\begin{array}{l} – 4\lambda z – 12\lambda ^2 + 2\lambda z = 0 \cdots (7) \\ 48\lambda ^2 – 16\lambda z = 36 \cdots (8) \end{array}} \right.\mathop \Rightarrow \limits^{(7)} 12\lambda ^2 + 2\lambda z = 0\Rightarrow 6\lambda + \lambda z = 0\Rightarrow \lambda=0\text{ or }=-6\lambda$若$x=0=y\mathop \Rightarrow \limits^{(5)} 0 = 36\ (\to\leftarrow)$若$z = – 6\lambda \Rightarrow (x,y,z) = ( – 4\lambda , – 4\lambda , – 6\lambda )\mathop \Rightarrow \limits^{(8)} 48\lambda ^2 + 96\lambda ^2 = 36 \Rightarrow \lambda = \pm \frac{1}{2}\Rightarrow (x,y,z) = ( – 2, – 2, – 3)\text{ or }(2,2,3)\because\ x,y,z>0\therefore\ (x,y,z) = (2,2,3) \Rightarrow \text{Max.}=2 \cdot 2 \cdot 3 = 12　∎$講解影片YouTube課程平台 此題考點Lagrange 乘數法 第二題 (12 分) Use the limits definition to show that$g'(0)$exists but$g'(0) \ne \mathop {\lim }\limits_{x \to 0} g'(x)$, where$g(x) = \left\{ {\begin{array}{c} {x^2 \sin \frac{1}{x},} & {{\text{if }}x \ne 0} \\ {0,} & {{\text{if }}x = 0}\end{array}} \right.$. 提示 此題要求用定義計算$x=0$時的微分，和用微分性質的結果作比較。 解答$\begin{array}{cl} g'(0) & =\mathop {\lim }\limits_{h \to 0} \frac{{g(0 + h) – g(0)}}{h} \\ & =\mathop {\lim }\limits_{h \to 0} \frac{{h^2 \sin \frac{1}{h} – 0}}{h} \\ & =\mathop {\lim }\limits_{h \to 0} h \cdot \sin \frac{1}{h} = 0 \end{array}$For$x\ne 0\begin{array}{cl} g'(x) & =2x \cdot \sin \frac{1}{x} + x^2 (\cos \frac{1}{x}) \cdot \frac{{ – 1}}{x} \\ & =2x\sin \frac{1}{x} – \cos \frac{1}{x} \end{array}\Rightarrow \mathop {\lim }\limits_{x \to 0} g'(x) = \mathop {\lim }\limits_{x \to 0} (2x \cdot \sin \frac{1}{x} – \cos \frac{1}{x})$D.N.E.$\Rightarrow g'(0) \ne \mathop {\lim }\limits_{x \to 0} g'(x)　∎$講解影片YouTube課程平台 此題考點導數與微分的概念 第三題 (12 分) Determine if the series converges or diverges 〔1〕$\sum\limits_{n = 0}^\infty {e^{ – n^2 } }$(6 分) 〔2〕$\sum\limits_{n = 1}^\infty {\sin \frac{1}{n}}$. (6 分) 提示 級數和的歛散性是必考的題目，因此八大審斂法需要很熟悉。第一題和高斯積分有關，可以配合積分審斂法；第二題則可以用$\mathop {\lim }\limits_{x\to 0} \frac{\sin x}{x}$的結論搭配極限比較審斂法。 解答 〔1〕 ◎ 高斯積分 Let${\rm{I}} = \int_0^\infty {e^{ – x^2 } } dx\begin{array}{cl} \rm{I}^2 & =\int_0^\infty {e^{ – y^2 } } dy\int_0^\infty {e^{ – x^2 } } dx \\ & =\int_0^\infty {\int_0^\infty {e^{ – x^2 } e^{ – y^2 } } dx} dy \\ & =\int_0^{\frac{\pi }{2}} {\int_0^\infty {e^{ – r^2 } } dr} d\theta\\&=\frac{\pi }{2} \cdot \frac{{\rm{1}}}{{\rm{2}}} = \frac{\pi }{4}  \end{array}\Rightarrow \int_0^\infty {e^{ – x^2 } } dx = \frac{{\sqrt \pi }}{2}\because\ \int_0^\infty {e^{ – x^2 } } dx = \frac{{\sqrt \pi }}{2}\therefore\ \sum\limits_{n = 1}^\infty {e^{ – n^2 } }\text{ converges. (by integral test)}　∎$(更嚴謹的做法請參考 A3, A4, A7 卷) 〔2〕$\begin{array}{cl} \because & \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{1}{n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{\theta \to 0^ + } \frac{{\sin \theta }}{\theta } = 1\ \text{Let }\theta=\frac{1}{n} \\ & \text{and }\sum\limits_{n = 1}^\infty {\frac{1}{n}}\ \text{diverges. }(\text{by }p\text{-series test}) \\ \therefore & \sum\limits_{n = 1}^\infty {\sin \frac{1}{n}}\ \text{diverges. (by limit comparison test)}　∎ \end{array}\$