401. The graph of \( f \) is shown below. Find the average value of \( f \) on the interval \( [-1,6] \).
答案： ( D )
Average value = \( \frac{1}{b-a}\int_a^b f(x)\,dx \). The integral is the net area; \( \frac{1}{7}\int_{-1}^6 f(x)\,dx = \frac{4}{7} + \frac{\pi}{14} \).
402. The velocity of a particle moving on a line is given by \( v(t) = 2t^2 - 14t - 5 \). Find the average velocity from \( t = 1 \) to \( t = 3 \).
答案： ( D )
\( \frac{1}{3-1}\int_1^3 (2t^2-14t-5)\,dt = \frac{1}{2}[\frac{2t^3}{3}-7t^2-5t]_1^3 = -\frac{73}{3} \).
403. What is the average value of \( y = 3\sin(2x) - 3\cos(2x) \) on the interval \( [-\frac{\pi}{2}, \frac{\pi}{6}] \)?
答案： ( C )
Average = \( \frac{1}{\pi/6-(-\pi/2)}\int_{-\pi/2}^{\pi/6} (3\sin(2x)-3\cos(2x))\,dx \); evaluates to \( -\frac{27}{8\pi} - \frac{9\sqrt{3}}{8\pi} \).
404. The graph of the acceleration function of a moving particle is shown below. On what intervals in \( [0,20] \) does the particle have a positive change in velocity, and what is the change in velocity over the entire interval [0, 20]?
答案： ( A )
Area under \( a(t) \) gives change in velocity. Positive on \( [5,15] \); total area = \( -\frac{25}{2} \) m/s.
405. The velocity of a particle is given by \( v(t) = 7\cos(\frac{1}{7}t^2 + 1) \) for \( t \geq 0 \). At \( t = 1 \) the position is 10. Using your calculator, find the position when the velocity is equal to 0 for the first time.
答案： ( D )
First zero of \( v \) at \( t = 2 \). Position = \( 10 + \int_1^2 7\cos(\frac{t^2}{7}+1)\,dt \approx 11.635 \) m.
406. A particle is moving along a straight line, and its acceleration is shown below. Assuming the particle started at rest, calculate the speed of the particle after 3 s.
答案： ( D )
\( v(3) = \int_0^3 a(t)\,dt \) = area under curve = \( -3 \) m/s. Speed = \( |v| = 3 \) m/s.
407. The graph of the velocity function of a moving car is shown below. In the interval \( [0,15] \), calculate (total distance of the car) - (total displacement of the car).
答案： ( B )
Displacement = \( \int_0^{15} v\,dt = 35 \) m; total distance = \( \int_0^{15} |v|\,dt = 65 \) m; difference = 30 m.
408. The velocity function of a moving particle is given by \( v(t) = t^2 - t - 2 \) for \( 0 \leq t \leq 6 \). Find (total distance traveled) - (total displacement).
答案： ( A )
Displacement = \( \int_0^6 v\,dt = 42 \); distance = \( \int_0^2 |v|\,dt + \int_2^6 |v|\,dt = \frac{146}{3} \); difference = \( \frac{20}{3} \) m.
409. The velocity function of a particle is \( v(t) = t^3 - 3t^2 - 2t + 4 \) (m/s). For \( 0 \leq t \leq 5 \) find (total displacement) - (total distance traveled).
答案： ( A )
Displacement = \( \frac{105}{4} \); distance (piecewise over zeros) = 38.75 m; displacement - distance = \( -12.5 \) m.
410. The acceleration of a particle is \( a(t) = -3 \) with \( v_0 = 9 \) for \( 0 \leq t \leq 7 \) (m/s²). Find the total distance traveled.
答案： ( D )
\( v(t) = -3t + 9 \); zero at \( t = 3 \). Distance = \( \int_0^3 (-3t+9)\,dt + \int_3^7 (3t-9)\,dt = \frac{75}{2} \) m.
411. The velocity of a particle on a coordinate line is \( v(t) = 2\sin(3t) \) for \( -\pi \leq t \leq \pi \). Using a calculator, find the total distance traveled when the particle is moving to the right.
答案： ( B )
Integrate \( |v(t)| \) over intervals where \( v > 0 \); sum of areas \( \approx 4 \).
412. The marginal profit is given by \( P'(x) = 1000 - 0.04x \), where \( x \) is the number of units sold. How much profit should the company expect if it sells 20,000 units?
答案： ( A )
Profit = \( \int_0^{20000} (1000 - 0.04x)\,dx = 1000(20000) - 0.02(20000)^2 = 12{,}000{,}000 \).
413. The marginal cost of producing \( x \) units is \( C'(x) = 5 + 0.4x \). Find the cost of producing the first 100 pairs.
答案： ( A )
Cost = \( \int_0^{100} (5 + 0.4x)\,dx = 5(100) + 0.2(100)^2 = 2{,}500 \).
414. Graphs of the marginal cost and marginal revenue are shown below. Estimate the profit for the first 50 units.
答案： ( D )
Profit = revenue - cost = area under MR minus area under MC \( = 1000(50)(0.5) - 50(50)(0.5) = 23{,}750 \).
415. The temperature of a hot penny changes at a rate \( T'(t) = -15e^{-0.12t} \) for \( 0 \leq t \leq 5 \) min, with temperature in Fahrenheit. If the penny is initially 150°F, find the temperature to the nearest degree after 4 min.
答案： ( C )
Change in temp = \( \int_0^4 (-T'(t))\,dt = \int_0^4 15e^{-0.12t}\,dt \approx 47.65 \); temperature \( \approx 150 - 47.65 = 102 \)°F.
416. The rate of change in temperature from 6 a.m. to 10 p.m. is \( f(t) = 3\cos(\frac{t}{3}) \) °F, where \( t \) is hours after 6 a.m. If at 6 a.m. the temperature is 75°F, find the temperature at 1 p.m.
答案： ( B )
\( \int_0^7 3\cos(t/3)\,dt \approx 6.51 \); temperature at 1 p.m. = \( 75 + 6.51 = 81.51 \)°F.
417. The rate of change in air temperature at O'Hare is shown below. If the temperature is 60°F at 5 a.m., what number best approximates the temperature at 8 a.m.?
答案： ( C )
Net change from area under curve: \( 60 - 2(2)(0.5) + 1(1)(0.5) = 58.5 \)°F.
418. Oil is leaking at a rate \( f(t) = 12e^{-0.4t} \) oz/h, \( t \) in hours. How many ounces will have leaked after 12 hours?
答案： ( C )
\( \int_0^{12} 12e^{-0.4t}\,dt = \frac{12}{-0.4}[e^{-0.4t}]_0^{12} = 29.7531 \) oz.
419. Water is leaking from a tank at a rate \( f(t) \); the graph of \( f(t) \) is shown. What is the best approximation of the total amount of water leaked for \( 5 \leq t \leq 10 \)?
答案： ( A )
Area under the curve from \( t = 5 \) to \( t = 10 \); trapezoid approximation \( \approx 88 \).
420. A full water tank leaks at a rate \( f(t) = 7e^{-0.1t} + 5 \) gal/h, \( t \) in hours. After 12 hours the tank is half empty. How much water did the full tank contain?
答案： ( A )
Leaked in 12 h = \( \int_0^{12} (7e^{-0.1t}+5)\,dt \approx 108.916 \) gal; full tank = \( 2(108.916) \approx 217.833 \) gal.
421. Sales of BBQs increase at a rate \( f(t) = 30 + 20\ln(1+t) \) per week, \( t \) in weeks. Texas Depot sold no BBQs in January. How many BBQs will be sold by the end of the 7th week?
答案： ( D )
\( \int_0^7 (30+20\ln(1+t))\,dt \approx 403 \) BBQs.
422. A colony of bacteria grows at a rate \( f(t) = 100 + 250\ln(3+t) \), \( t \) in hours. How much will the population increase between the 6th and 8th hours?
答案： ( D )
\( \int_6^8 (100+250\ln(3+t))\,dt \approx 1{,}350 \).
423. The rate of hybrid car production 2001–2011 is shown below. Which number best approximates the increase in production between 2008 and 2011?
答案： ( C )
Area under the rate curve from 2008 to 2011 \( \approx 3(65000) + \frac{1}{2}(3)(30000) = 240{,}000 \).
424. Stock price is approximated by \( f(x) = \frac{x^2}{4} + 2 \), where \( x \) = hours after 9 a.m. (0 to 6). Is there a time when the value equals its average for the 6 hours? If yes, when?
答案： ( A )
MVT for integrals: \( \frac{1}{6}\int_0^6 f(x)\,dx = f(c) \) gives \( c = 2\sqrt{3} \approx 3.46 \) h after 9 a.m. → 12:28 p.m.
425. Given \( f(x) = \sqrt{2x-3} \), verify the Mean Value Theorem for integrals on \( [2,8] \) and find \( c \).
答案： ( C )
\( \frac{1}{6}\int_2^8 \sqrt{2x-3}\,dx = f(c) \); \( c \approx 4.75 \).
426. Plutonium has a half-life of 8,645 years. If there are initially 20 g, how many grams are left after 1,000 years?
答案： ( B )
\( y = 20 e^{kt} \), \( k = -\frac{\ln 2}{8645} \); \( y(1000) = 20 e^{1000k} \approx 18 \) g.
427. Bacteria increase at a rate proportional to the amount present. There are 200 after 1 day and 600 after 3 days. How many after 7 days?
答案： ( A )
\( y = y_0 e^{kt} \); from \( 200 = y_0 e^k \), \( 600 = y_0 e^{3k} \) get \( e^{2k}=3 \); \( y(7) \approx 5{,}400 \).
428. Coffee temperature decreases by \( \frac{dy}{dt} = ky \) with \( t \) in minutes. After 5 min the temperature decreases by 70%. Then \( k = \)? (most complete answer)
答案： ( D )
\( 0.3y_0 = y_0 e^{5k} \) ⇒ \( k = \frac{\ln 0.3}{5} = \frac{\ln(3/10)}{5} = \frac{\ln 3 - \ln 10}{5} \); B and C are equivalent.
429. The wolf population in Yosemite increased by 11% between 2002 and 2011. What is the constant of proportionality?
答案： ( D )
\( 1.11 = e^{9k} \) ⇒ \( k = \frac{\ln 1.11}{9} \approx 0.0116 \).
430. For an item starting from rest, which expression(s) yield the velocity at the end of an interval? I. \( \frac{ds}{dt} \) II. \( \frac{dv}{dt} \) III. \( \int_0^t a(x)\,dx \) IV. \( \int_0^t v(x)\,dx \)
答案： ( C )
Velocity = \( \frac{ds}{dt} \); from rest, \( v(t) = \int_0^t a(x)\,dx \).
431. If \( \frac{dy}{dx} = -x\cos(x^2) \) and \( y = 2 \) when \( x = 0 \), a solution is:
答案： ( D )
Separate and integrate: \( dy = -x\cos(x^2)\,dx \); with \( u = x^2 \), \( y = -\frac{1}{2}\sin(x^2)+C \); \( y(0)=2 \) ⇒ \( C=2 \).
432. If \( \frac{dy}{dx} = 5x^4 y^2 \) and \( y = 1 \) when \( x = 1 \), determine \( y \) when \( x = -1 \).
答案： ( C )
\( \frac{1}{y^2}\,dy = 5x^4\,dx \) ⇒ \( -\frac{1}{y} = x^5 + C \); \( y(1)=1 \) ⇒ \( C=-2 \); \( y(-1) = \frac{1}{3} \).
433. If \( \frac{d^2 y}{dx^2} = 4x - 5 \) with \( x=0 \), \( y'=3 \) and \( y=4 \), find a solution.
答案： ( A )
Integrate twice: \( y' = 2x^2 - 5x + C_1 \), \( y'(0)=3 \) ⇒ \( C_1=3 \); \( y = \frac{2}{3}x^3 - \frac{5}{2}x^2 + 3x + C_2 \), \( y(0)=4 \) ⇒ \( C_2=4 \).
434. The figure shows a slope field for one of the differential equations below. Identify the equation.
答案： ( B )
When \( x=0 \) the slope is 0; when \( x,y > 0 \) slope is negative. Only (B) \( \frac{dy}{dx} = -xy \) matches.
435. The figure shows a slope field for one of the differential equations below. Identify the equation.
答案： ( A )
Same slope in each vertical column; positive for \( x>0 \), negative for \( x<0 \) ⇒ \( \frac{dy}{dx} = x \).
436. BC: What family of functions has a slope field given by \( \frac{dy}{dx} = \frac{x}{y} \)?
答案： ( C )
\( y\,dy = x\,dx \) ⇒ \( \frac{y^2}{2} = \frac{x^2}{2} + C \) ⇒ \( x^2 - y^2 = \) constant (hyperbolas).
437. BC: If \( \frac{dy}{dx} = y - x \), identify the slope field. (Figure 1, Figure 2, Figure 3, or none)
答案： ( B )
Table of slopes for \( \frac{dy}{dx} = y - x \) matches Figure 2.
438. BC: The rabbit population was 17 in 2001 and 35 in 2008. Using a logistic model with carrying capacity 75, predict the population in 2013.
答案： ( D )
Logistic \( P(t) = \frac{K}{Ae^{-kt}+1} \); fit \( P(0)=17 \), \( P(7)=35 \); \( P(12) \approx 49 \).
439. BC: Spread of beetles through 60 trees: \( \frac{dP}{dt} = 0.71P(1-\frac{P}{60}) \), one tree infected at day 0. When will half the trees be infected?
答案： ( C )
Logistic solution; solve \( P(t) = 30 \) ⇒ \( t \approx 6 \) days.
440. BC: A school has 1,532 students. A rumor: 7 have heard on Monday, 84 on Tuesday. How many will have heard by Thursday?
答案： ( A )
Logistic model with \( K=1532 \), \( P(0)=7 \), \( P(1)=84 \); \( P(3) \approx 1{,}384 \) students.
441. BC: A dorm of 300 students has a measles outbreak. 4 are diagnosed on day 0, 17 after 7 days. Use a logistic model to predict the number infected after 14 days.
答案： ( D )
Logistic \( P(t) = \frac{300}{74e^{-kt}+1} \) with \( P(0)=4 \), \( P(7)=17 \); solve for \( k \); \( P(14) \approx 64 \).
442. BC: Given \( \frac{dy}{dt} = 2\sin(4\pi t) \) and \( y(1) = 2 \), approximate \( y(3) \) using five steps.
答案： ( D )
Euler with \( \Delta t = 0.4 \); or \( y(3) = y(1) + \int_1^3 2\sin(4\pi t)\,dt = 2 \).
443. BC: Use Euler's method with step size 0.2 to compute \( y(1) \) if \( \frac{dy}{dx} + 4x^3 y = 2x^3 \) and \( y(0) = 2 \).
答案： ( B )
\( \frac{dy}{dx} = 2x^3(1-2y) \); Euler table gives \( y(1) \approx 1.19 \).
444. BC: Approximate \( P(4) \) given \( \frac{dP}{dt} = -0.2P(1-\frac{3P}{5}) \) with \( P(2) = 3 \). Use three steps.
答案： ( A )
Euler with \( \Delta t = 2/3 \); \( P(4) \approx 4.4 \).
445. BC: The velocity of a ball thrown from a 50-ft cliff is shown below. Use Euler's method with step size 0.5 to approximate the distance from the ground after 2 s.
答案： ( C )
Euler on \( v \) from initial height 50; after 2 s distance from ground \( \approx 38.5 \) ft.
446A. 446. The rate of change in temperature of a greenhouse from 7 p.m. to 7 a.m. is \( f(t) = -3\sin(\frac{t}{3}) \) °F, \( t \) = hours after 7 p.m. (A) If at 7 p.m. the temperature is 105°F, find the temperature at 2 a.m.
\( F(7) = 105 + \int_0^7 (-3\sin(t/3))\,dt = 105 + 9[\cos(t/3)]_0^7 \approx 89.78 \)°F.
446B. 446. (B) Write an integral expression for the temperature of the greenhouse at time \( t \) (between 7 p.m. and 7 a.m.).
\( F(t) = 105 + \int_0^t (-3\sin(x/3))\,dx \).
446C. 446. (C) Find the average change in temperature between 7 p.m. and 7 a.m. to the nearest tenth of a degree.
Average = \( \frac{1}{12}\int_0^{12} (-3\sin(t/3))\,dt \approx -1.24 \)°F per hour.
446D. 446. (D) Is there a time during the night when the rate of change equals the average from part (c)? If yes, state the time. Justify.
By MVT for integrals, yes. Solve \( -3\sin(t/3) = -1.24 \) ⇒ \( t \approx 1.28 \) h after 7 p.m. → about 8:17 p.m.
447A. 447. Consider \( \frac{dy}{dx} = \frac{4xy}{5} \). (A) On the axes provided, sketch a slope field at the points indicated.
Set up a table of \( \frac{dy}{dx} = \frac{4xy}{5} \) at the given points and sketch short tangent segments.
447B. 447. (B) Let \( y = f(x) \) be the solution with \( f(0) = 5 \). Use Euler's method with step 0.1 to approximate \( f(0.3) \).
\( y_0=5 \), \( y_1=5.04 \), \( y_2=5.1206 \); \( f(0.3) \approx 5.1206 \).
447C. 447. (C) Find the particular solution \( y = f(x) \) with \( f(0) = 5 \).
\( \frac{1}{y}\,dy = \frac{4}{5}x\,dx \) ⇒ \( \ln|y| = \frac{2}{5}x^2 + C \); \( y = 5e^{\frac{2}{5}x^2} \).
447D. 447. (D) Use your solution in (c) to find \( f(0.3) \).
\( f(0.3) = 5e^{\frac{2}{5}(0.09)} = 5e^{0.036} \approx 5.1833 \).
448A. 448. The slope of \( f \) at \( (x,y) \) is \( \frac{2y}{3x^2} \); the point \( (3,4) \) is on the graph. (A) Write an equation of the tangent line at \( x = 3 \).
Slope at \( (3,4) \) is \( \frac{8}{27} \); tangent: \( y - 4 = \frac{8}{27}(x-3) \) or \( y = \frac{8}{27}x + \frac{28}{9} \).
448B. 448. (B) Use the tangent line to approximate \( f(5) \). Is this an overestimate or underestimate? Justify.
\( f(5) \approx \frac{124}{27} \). Overestimate because the graph is concave down (second derivative negative).
448C. 448. (C) Solve the separable differential equation \( \frac{dy}{dx} = \frac{2y}{3x^2} \) with initial condition \( f(3) = 4 \).
\( \frac{dy}{y} = \frac{2}{3}x^{-2}\,dx \) ⇒ \( \ln|y| = -\frac{2}{3x} + C \); \( y = ce^{-2/(3x)} \); \( f(3)=4 \) ⇒ \( y \approx 4.9954\,e^{-2/(3x)} \).
448D. 448. (D) Use the solution in (c) to find \( f(5) \).
\( f(5) = 4.9954\,e^{-2/15} \approx 4.37 \).
449A. 449. A water barrel contains 200 gal and leaks at \( 15\sin(\frac{\pi t}{30}) \) gal/min, \( 0 \leq t \leq 15 \). (A) How much water, to the nearest gallon, leaked out after \( t = 7 \) min?
\( \int_0^7 15\sin(\frac{\pi t}{30})\,dt = \frac{450}{\pi}(1-\cos(\frac{7\pi}{30})) \approx 36.79 \) → 37 gal.
449B. 449. (B) What is the average amount of water leaked per minute from \( t = 0 \) to \( t = 7 \) to the nearest gallon?
Average = \( \frac{1}{7}\int_0^7 15\sin(\frac{\pi t}{30})\,dt \approx 5.26 \) → 5 gal/min.
449C. 449. (C) Write an expression for \( f(t) \) = total amount of water in the barrel at time \( t \).
\( f(t) = 200 - \int_0^t 15\sin(\frac{\pi x}{30})\,dx = 200 + \frac{450}{\pi}(\cos(\frac{\pi t}{30})-1) \).
449D. 449. (D) At what \( t \) to the nearest minute will there be 100 gal remaining?
Solve \( 100 = 200 - \int_0^a 15\sin(\frac{\pi t}{30})\,dt \) ⇒ \( \cos(\frac{\pi a}{30}) \approx 0.302 \) ⇒ \( a \approx 12 \) min.
450A. BC 450. \( \frac{dP}{dt} = \frac{1}{7}P(1-\frac{P}{21}) \). (A) If \( P(0) = 15 \), what is \( \lim_{t \to \infty} P(t) \)?
Logistic with carrying capacity 21; \( \lim_{t \to \infty} P(t) = 21 \).
450B. 450. (B) For what value of \( P \) is the population growing fastest?
Growth rate \( \frac{dP}{dt} \) is maximum when \( P = \frac{K}{2} = \frac{21}{2} = 10.5 \).
450C. 450. (C) A population \( Q \) satisfies \( \frac{dQ}{dt} = \frac{1}{7}Q(1-\frac{t}{21}) \) with \( Q(0) = 10 \). Find \( Q(t) \).
Separate: \( \frac{dQ}{Q} = \frac{1}{7}(1-\frac{t}{21})\,dt \); \( \ln|Q| = \frac{t}{7} - \frac{t^2}{294} + C \); \( Q(t) = 10e^{t/7 - t^2/294} \).
450D. 450. (D) For \( Q \) in (c), what is \( \lim_{t \to \infty} Q(t) \)?
Exponent \( \frac{t}{7} - \frac{t^2}{294} \to -\infty \) as \( t \to \infty \); so \( \lim Q(t) = 0 \).