The function $f$ is continuous for $- 4 \leq x \leq 4$. The graph of $f$ shown above consists of five line segments. What is the average value of $f$ on the interval $- 4 \leq x \leq 4$ ?
(A) $\frac { 1 } { 8 }$
(B) $\frac { 3 } { 16 }$
(C) $\frac { 15 } { 16 }$
(D) $\frac { 3 } { 2 }$

21. The average value of $\frac { 1 } { x }$ on the closed interval $[ 1,3 ]$ is
(A) $\frac { 1 } { 2 }$
(B) $\frac { 2 } { 3 }$
(C) $\frac { \ln 2 } { 2 }$
(D) $\frac { \ln 3 } { 2 }$
(E) $\ln 3$

5. Let $R$ be the region in the first quadrant under the graph of $y = \frac { x } { x ^ { 2 } + 2 }$ for $0 \leqq x \leqq \sqrt { 6 }$.
(a) Find the area of $R$.
(b) If the line $\mathrm { x } = k$ divides $R$ into two regions of equal area, what is the value of $k$ ?
(c) What is the average value of $y = \frac { x } { x ^ { 2 } + 2 }$ on the interval $0 \leqq x \leqq \sqrt { 6 }$ ?
Solution Distribution of Points
(a) $A = \int _ { 0 } ^ { \sqrt { 6 } } \frac { x } { x ^ { 2 } + 2 } d x$
$$\begin{aligned} & = \left. \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 2 \right) \right| _ { 0 } ^ { \sqrt { 6 } } \\ & = \frac { 1 } { 2 } \ln 8 - \frac { 1 } { 2 } \ln 2 = \ln 2 \end{aligned}$$
(b) $\frac { 1 } { 2 } \ln 2 = \int _ { 0 } ^ { k } \frac { x } { x ^ { 2 } + 2 } d x$
$$\begin{aligned} & = \left. \frac { 1 } { 2 } \ln \left( x ^ { 2 } + 2 \right) \right| _ { 0 } ^ { k } \\ & = \frac { 1 } { 2 } \ln \left( k ^ { 2 } + 2 \right) - \frac { 1 } { 2 } \ln 2 \end{aligned}$$
$$\begin{aligned} & \therefore \frac { 1 } { 2 } \ln \left( k ^ { 2 } + 2 \right) = \frac { 1 } { 2 } \ln 2 + \frac { 1 } { 2 } \ln 2 \\ & \quad \text { or } \ln \left( k ^ { 2 } + 2 \right) = \ln 4 \\ & \therefore k ^ { 2 } + 2 = 4 \text { and } k = \sqrt { 2 } \end{aligned}$$
(c) Average value $= \frac { 1 } { \sqrt { 6 } - 0 } \int _ { 0 } ^ { \sqrt { 6 } } \frac { \mathrm { x } } { \mathrm { x } ^ { 2 } + 2 } \mathrm { dx }$
$$= \frac { 1 } { \sqrt { 6 } } \ln 2$$
(a) 1: for correct integral $3 : \left\{ \begin{array} { l } 1 : \text { for antiderivative } \\ 1 : \text { for evaluation } \end{array} \right.$
(b) $\quad \begin{cases} 1 : & \text { for a correct equation } \\ & \text { involving integral(s) } \\ 1 : & \text { for antiderivative } \\ 1 : & \text { for evaluation of integral(s) } \end{cases}$
1: for finding the value of $k$
(c) $2 : \left\{ \begin{array} { l } 1 : \text { for correct integral } \\ 1 : \text { for evaluation } \end{array} \right.$

The temperature outside a house during a 24-hour period is given by $$F(t) = 80 - 10\cos\left(\frac{\pi t}{12}\right), \quad 0 \leq t \leq 24,$$ where $F(t)$ is measured in degrees Fahrenheit and $t$ is measured in hours.
(a) Sketch the graph of $F$ on the grid provided.
(b) Find the average temperature, to the nearest degree Fahrenheit, between $t = 6$ and $t = 14$.
(c) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of $t$ was the air conditioner cooling the house?
(d) The cost of cooling the house accumulates at the rate of $\$0.05$ per hour for each degree the outside temperature exceeds 78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 24-hour period?

Let $f$ be the function defined by $$f(x) = \begin{cases} \sqrt{x+1} & \text{for } 0 \leq x \leq 3 \\ 5 - x & \text{for } 3 < x \leq 5. \end{cases}$$ (a) Is $f$ continuous at $x = 3$? Explain why or why not.
(b) Find the average value of $f(x)$ on the closed interval $0 \leq x \leq 5$.
(c) Suppose the function $g$ is defined by $$g(x) = \begin{cases} k\sqrt{x+1} & \text{for } 0 \leq x \leq 3 \\ mx + 2 & \text{for } 3 < x \leq 5, \end{cases}$$ where $k$ and $m$ are constants. If $g$ is differentiable at $x = 3$, what are the values of $k$ and $m$?

Let $R$ be the region in the first quadrant under the graph of $y = \frac { x } { x ^ { 2 } + 2 }$ for $0 \leqq x \leqq$ (a) Find the area of $R \cdot \left( - x - \alpha , \frac { 1 } { 1 } \right)$ (b) If the line $x = k$ divides $R$ into two regions of equal area, what is the value of $k$ ? (c) What is the average value of $y = \frac { x } { x ^ { 2 } + 2 }$ on the interval $0 \leqq x \leqq \sqrt { 6 }$ ?

An invasive species of plant appears in a fruit grove at time $t = 0$ and begins to spread. The function $C$ defined by $C ( t ) = 7.6 \arctan ( 0.2 t )$ models the number of acres in the fruit grove affected by the species $t$ weeks after the species appears. It can be shown that $C ^ { \prime } ( t ) = \frac { 38 } { 25 + t ^ { 2 } }$.
(Note: Your calculator should be in radian mode.)
A. Find the average number of acres affected by the invasive species from time $t = 0$ to time $t = 4$ weeks. Show the setup for your calculations.
B. Find the time $t$ when the instantaneous rate of change of $C$ equals the average rate of change of $C$ over the time interval $0 \leq t \leq 4$. Show the setup for your calculations.
C. Assume that the invasive species continues to spread according to the given model for all times $t > 0$. Write a limit expression that describes the end behavior of the rate of change in the number of acres affected by the species. Evaluate this limit expression.
D. At time $t = 4$ weeks after the invasive species appears in the fruit grove, measures are taken to counter the spread of the species. The function $A$, defined by $A ( t ) = C ( t ) - \int _ { 4 } ^ { t } 0.1 \cdot \ln ( x ) d x$, models the number of acres affected by the species over the time interval $4 \leq t \leq 36$. At what time $t$, for $4 \leq t \leq 36$, does $A$ attain its maximum value? Justify your answer.

The temperature outside a house during a 24-hour period is given by $$F(t) = 80 - 10\cos\left(\frac{\pi t}{12}\right), \quad 0 \leq t \leq 24,$$ where $F(t)$ is measured in degrees Fahrenheit and $t$ is measured in hours.
(a) Sketch the graph of $F$ on the grid provided.
(b) Find the average temperature, to the nearest degree Fahrenheit, between $t = 6$ and $t = 14$.
(c) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of $t$ was the air conditioner cooling the house?
(d) The cost of cooling the house accumulates at the rate of $\$0.05$ per hour for each degree the outside temperature exceeds 78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 24-hour period?

What is the average value of $y = \sqrt { \cos x }$ on the interval $0 \leq x \leq \frac { \pi } { 2 }$ ?
(A) - 0.637
(B) 0.500
(C) 0.763
(D) 1.198
(E) 1.882

123- What is the mean value of $f(x) = \dfrac{x^2 - 2}{x^2}$ on the interval $[2, 4]$?
(1) $\dfrac{5}{8}$ (2) $\dfrac{11}{16}$ (3) $\dfrac{3}{4}$ (4) $\dfrac{7}{8}$

Let $f : [0,2] \rightarrow \mathbb{R}$ be a continuous function such that $$\frac{1}{2}\int_0^2 f(x)\,dx < f(2)$$ Then which of the following statements must be true?
(A) $f$ must be strictly increasing.
(B) $f$ must attain a maximum value at $x = 2$.
(C) $f$ cannot have a minimum at $x = 2$.
(D) None of the above.

7. A resistor with resistance $R$ is traversed by a current varying in time with intensity $I ( t ) = I _ { 0 } \frac { a } { t }$, with $t > 0$ and the positive constants $I _ { 0 }$ and $a$ expressed, respectively, in amperes and in seconds. Knowing that the power dissipated in the resistor due to the Joule effect is $P ( t ) = R I ^ { 2 } ( t )$, determine its average value on the interval $[ 2 a ; 3 a ]$.

24. $\quad \int _ { 0 } ^ { \pi / 2 } \sin x d x$ is equal to
(A) $\frac { \pi } { 8 } ( 1 + \sqrt { 2 } )$
(B) $\frac { \pi } { 4 } ( 1 + \sqrt { 2 } )$
(C) $\frac { \pi } { 8 \sqrt { 2 } }$
(D) $\frac { \pi } { 4 \sqrt { 2 } }$
Sol. (A)
$$\begin{aligned} & \int _ { 0 } ^ { \pi / 2 } \sin x d x = \frac { \frac { \pi } { 2 } + 0 } { 4 } \left( \sin ( 0 ) + \sin \left( \frac { \pi } { 2 } \right) + 2 \sin \left( \frac { 0 + \frac { \pi } { 2 } } { 2 } \right) \right) \\ & = \frac { \pi } { 8 } ( 1 + \sqrt { 2 } ) \end{aligned}$$

1. Data could not be retrieved.
2. If $\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) < 0 \forall \mathrm { x } \in ( \mathrm { a } , \mathrm { b } )$ and c is a point such that $\mathrm { a } < \mathrm { c } < \mathrm { b }$, and (c, $\left. \mathrm { f } ( \mathrm { c } ) \right)$ is the point lying on the curve for which $\mathrm { F } ( \mathrm { c } )$ is maximum, then $\mathrm { f } ^ { \prime } ( \mathrm { c } )$ is equal to
   (A) $\frac { f ( b ) - f ( a ) } { b - a }$
   (B) $\frac { 2 ( f ( b ) - f ( a ) ) } { b - a }$
   (C) $\frac { 2 f ( b ) - f ( a ) } { 2 b - a }$
   (D) 0

Sol. (A)
$$\begin{aligned} & \left( F ^ { \prime } ( c ) = ( b - a ) f ^ { \prime } ( c ) + f ( a ) - f ( b ) \right. \\ & F ^ { \prime \prime } ( c ) = f ^ { \prime \prime } ( c ) ( b - a ) < 0 \\ & \Rightarrow F ^ { \prime } ( c ) = 0 \Rightarrow f ^ { \prime } ( c ) = \frac { f ( b ) - f ( a ) } { b - a } \end{aligned}$$

Comprehension III

Let ABCD be a square of side length 2 units. $\mathrm { C } _ { 2 }$ is the circle through vertices $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D }$ and $\mathrm { C } _ { 1 }$ is the circle touching all the sides of the square ABCD . L is a line through A .