[78%] AB calc cumulative exam 2 electric boogaloo

The area bounded by f(x) = 1 - x^2 and g(x) = x^2(x^2 -1) is 4/15 16/15 4/3 8/5

8/5

integrate (10^x - 1/sqrt(x))dx = 10^x - sqrtx + c (10^x)/ln10 - sqrtx + c 10^x - 2sqrtx + c (10^x)/ln10 - 2sqrtx + C

(10^x)/ln10 - 2sqrtx + C

integral from 0 to 4 (|x-3| - 1)dx 0 1 3 4

1

let v(t) = -1/2 t(t-2)(t-8) represetn an object's velocity at time t seconds. the total distance the object travels in the first 6 seconds is 24 54 63.33333333333333 94.666666666666666666666666pain666666666

54

The rate at which the population P, of a village is increasing is given by dP/dt = 0.011P where t is time, in years. If the village's population is currently 5000 people, how many years will it take for the village's population to double? 55 63 91 182

63

region R is bounded by f(x) = x^2 and g(x) = 3x. What is the volume of the solid formed when region R is revolved about the y axis? 3/2 pi 27/10 pi 7/2 pi 27/2 pi

7/2 pi you have to change things into in terms of x, so f(x) would become sqrty and g(x) would become y/3. then integrate that based on y.

the area of the region bounded by f(x) = -2 cuberoot(x-2) and g(x) = 1- 1/2 x is 0 8 16 80

8

which statement describes the value of the integral integrate 0 to 4 f(x)dx, where f(x) = 12-3x^2, 0<x<=2 and -4, 2<x<=4 8 12 24 DNE

8 integrate 12-3x^2 from 0 to 2, get 16 integrate -4 from 4 to 2, get -8 16 + (-8) = 8

Region R is bounded by y= 4-x^2 and the x axis. A solid has base R, and cross sections perpendicular to the y-axis are semicircles with the diameter located in the base. The volume of this solid is 16 pi 8 pi 4 pi pi

8 pi

Use the table that shows the rate of which vehicles pass through a toll booth from 6:00 am to 12:00 pm time - - vehicles per hour 6:00 am - - 72 6:30 am - - 90 7:00 - - 126 7:30 - - 186 8:00 - - 218 8:30 - - 180 9:00 - - 166 9:30 - - 150 10:00 - - 146 10:30 - - 126 11:00 - - 168 11:30 - - 182 12:00 - - ??? using a left riemmann sum with 6 equal subintervals, what is the total number of vehicles to pass through the toll booth from 6:00 am to 12:00 pm 896 914 918 1004

896 (add them up from 6:00, skip the xx:30s, and ignore what 12:00 is)

When the region bounded by y = 3e^2x and the x axis on the interval [-2,0] is revolved about the x acis. the volume of the resulting solid is found usin which of the following integrals? v = pi integral 0 to 3 9e^(4x) dx v = pi integral 0 to 3 3e^(4x) dx v = pi integral -2 to 0 9e^(4x) dx v = pi integral -2 to 0 3e^(4x) dx

v = pi integral -2 to 0 9e^(4x) dx

d/dx integrade from e to e^x (ln t)dt = x xe^x x-1 xe^x -1

xe^x

The equation dy/dt = -0.00012y gives the rate at which the amount, y, in milligrams, of a certain radioactive material changes over time, t , in years. if there are originally 10 milligras of the radioactive material, which fnction models the amount of radioactive material over time y = 10e^(-0.00012t) y = 10-e^(0.00012t) y = 10(1/2)^(-0.0012t) y = 10-(1/2)^(0.00012t)

y = 10e^(-0.00012t)

the solution of dy/dx = (2sqrty)/x passing through the point -1, 4 is y= ln^2 |x| + 2 ln^2 |x| + 4 (ln |x| + 2)^2 (ln |x| + 4)^2

(ln |x| + 2)^2 plug in y = the answer and calculate dy/dx to see if it matches

integral of 5x/e^(5x^2) dx = - 1/2e^(5x^2) + C 1/2e^(5x^2) + C - 2/e^(5x^2) + C 2/e^(5x^2) + C

- 1/2e^(5x^2) + C

The area under the curve is f(x) = 1/x +2 over the interval [1,e] is 1 3 -1+e^-1 -1 + 2e

-1 + 2e

integral of 1 to 2 of x/(1-4x^2) dx = -1/8 ln5 -1/8 ln2 1/8 ln2 1/8 ln5

-1/8 ln5

the function r(t) = 1000 pi cos (pi/2 t) gives the rate of change in the deer population in a certain area after t months. The net change in the deer population betwen t = 0 and t = 3 is -2000pi -2000 2000 2000pi

-2000

if f and g are functions such that integral from 0 to 2 f(x)dx = 2 and integral from 0 to 2 (f(x) - 2 g(x))dx = 8, what is the value of integral from 0 to 2 g(x) dx? -12 -3 3 12

-3 integral from 0 to 2 (f(x) - 2 g(x))dx = 8 = 2 - 2 integral from 0 to 2 g(x) dx = 8 = 2 integral from 0 to 2 g(x) dx = -6 = integral from 0 to 2 g(x) dx = -3

if integral of 1 to -1 f(x)dx = 3, then integral of -1 to 1 (f(x)-1)dx is -5 -4 1 2

-5 f(-1) - f(1) = 3 split integral of -1 to 1 (f(x)-1)dx into: = -(f(-1) - f(1)) - (1 - (-1)) = -3 - (2) = -5

let integral from -2 to 2 f'(x)dx = 6. if f(-2) = -2, then f(2) = -8 -4 4 12

-8

limit as x approaches infinity of ((20x^4 -1)/(4e^x -1)) 0 5 20 120

0 (graph it)

to the nearest thousandth, what is the area of the region bounded by y = e^x and x = (y-1)^2? 0.089 1.188 2.880 4.265

0.089

an object's velocity is given by v(t) = e^(1-t) -1 over the interval [0,2]. To the nearest thousandth, what is the object's displacement over this interval? 0.350 0.543 1.086 2.350

0.350

integrate 0 to 4 (|x - 3| - 1)dx = 0 1 3 4

1

region r is bounded by y=1/2 cos (pi x)/(2) and the x axis over the interval [-1,1]. a solid has base r, and cross sections perpendicular to the x axis are righ isoceles triangles with one leg located in the base. the volume of this solid is 1/16 1/8 pi/8 pi/4

1/16 do the pi 1/4 and then square thingie and slap that other equation in and then integrate over interval

an approximation of the area under the cirve of f(x) = cos(pi x) + 2^x over the interval [0,3] using a trapezoidal rule with 3 equal subintervals is 7.5 10.5 15 21

10.5

the rate, in liters per minute, at which water is being pumped out of an underground tank is given by the function r(t) = t^2 +1 for 0 <= t <=3. The total amount of water pumped in the first 3 minutes is 4 9 10 12

12 liters

the midpoint approximation of the area under the curve f(x) = 2x(x-4)(x-8) over the interval [0,4] with 4 subintervals is 111 120 132 160

120

use the 3 functions shown 1) f(x) = x^2 2) f(x) = sqrtx 3) f(x) = lnx for which of the functions, if any, is a trapezoidal approximation for the area under the cirve y = f(x) over the interval [1,4] with 3 equal subintervals less than the midpoint approximation? none of the functions 1 only 2 and 3 only all of the functions

2 and 3 only except i think i bullshitted this question so idk

use this graph of the velocity, v(t), in kilometers per hour, of a train over time t, in hours <<for 1.5 on the x axis, it's at 100. for the next 0.5, smoothly goes from 100 to 60. for the last 1.0 on the x axis, it's at 60>> 240 250 260 270

250

use the 3 functions shown 1) f(x) = 0.1(x-1) 2) f(x) = sqrt(x-1) 3) f(x) = ln(x-1) which inequality shows the growth rates from least to greatest as x approaches infinity 1 < 2 < 3 1 < 3 < 2 3 < 1 < 2 3 < 2 < 1

3 < 2 < 1 this was from graphing it, not looking at the derivatives, so it could be wrong

if f(x) = (ln3)(3^x), for what value of C, to the nearest thousandth, is f(c) equal to the average value of f over the interval [1, 4] 2.880 2.966 3.249 3.880

3.880

use this table of values that gives the rate at which people enter a line for concert tickets (wow imagine getting in line for something and breaking social distancing) t minutes ater the first person starts forming a line minutes - - people per minute 0 - - 1 10 - - 2 20 - - 5 30 - - 2 40 - - 9 50 - - 5 60 - - 13 based on a trapezoidal approximation, what is the total number of people in line at t = 60 minutses when the ticket counter finally opens? 240 300 360 420

300

when the region bounded by y = x^3 -1, the y axis, and the line y = 7 is revolved around the y axis, the volume of the resulting solid is 36.577 38.597 58.434 60.319

38.597 graph first, integrate between intersections, and then do ye olden disk method

integral of (4/x -1) dx = 4 ln x - 1 + C ln(4x) - 1 + C 4 ln x - x + C ln(4x) - x + C

4 ln x - x + C

use this graph of f consisting of a quarter circle and two line segments quarter circle spans 6 on y axis to 2 on y axis for 4 grid squares. then directly up in diagonal for 2 grid squares up to 6 again, where it stays as a horizontal line for 4 grid squares based on the graph, integrating 0 to 10 f(x) dx = 20 + pi 20 + 4pi 40 + pi 40 + 4 pi

40 + 4 pi

let f be a continuous function whose graph passes through points (4,0), (5,1), (8,2), (13,3) and (20,4). a trapezoidal approximation of the area under the graph of y = f(x) over the interval [4,20] with 4 ssubintervals is 32 38 42 50

42

The acceleration of a vehicle, in mph per second, is given by a(t) = -6. if the velocity of the vehicle is 60 mph at t = 2 seconds, the velocity at 5 seonds is 36 48 54 72

48 mph

Region R is bounded by f(x) = sqrt(2-x)^2 . What is the volume, to the nearest thousandth, of the solid formed when region r is revolved around the line y = -2? 0.404 5.131 6.687 15.184

5.131

let dy/dx -x^2y^2. which graph shows the particular solution of this differential equation passing through (-1, 1)? << a whole bunch of images of slope fields>

A (weird horizontl hiccup at y = 1.5 but it passes through (-1, 1)

let dy/dx = 2xy^(-2). the particular solution to this differential equation, which intersects the y axis at 3, is y = cuberoot (3x^2 -27) cuberoot (3x^2 + 3) cuberoot (3x^2 + 9) cuberoot (3x^2 + 27)

cuberoot (3x^2 + 27)

if dy/dx = -2(x-y), which slope field can be used to determine graphical solutions for y?

diagonal going up from left to right (B)intersected at x=1

Over the interval [0,2], the area under the curve y = f(x) with n equal subdivisions can be expressed as limit as n approaches infinity sum n, k = 1 2(Lk)^2 * 1/n, where Lk is the left endpoint of the kth subdivision. As a definite integral, the limit can be expressed as integral from 0 to 2 1/x dx integral from 0 to 2 2/x dx integral from 0 to 2 x^2 dx integral from 0 to 2 2x^2 dx

integral from 0 to 2 x^2 dx (C)

let r(t) represent the rate, in degrees Fahrenheit per hour, at which the temperature is changing t hours after 6 am. which definite integral represents the net change in temperature, in degrees Fahrenheit, between 9 am and 12 pm? integral from 3 to 6 r(t)dt integral from 9 to 12 r(t)dt integral from 3 to 6 r'(t)dt integral from 9 to 12 r'(t)dt

integral from 9 to 12 r(t)dt

Use this diagram (not included in this quizlet sorry) to answer the following question. <<its a pyramid on its side with sides of its square base of 4 cm>> A regualr square pyramid has a base with lengths of 4 cm and a heigh of 4 cm. Which definite integral represents the volume of this pyramid? integral of 0 to 4 x^2 dx integral of 0 to 4 1/2 x^2 dx integral of 0 to 4 1/3 x^2 dx integral of 0 to 4 1/4 x^2 dx

integral of 0 to 4 x^2 dx

the interval [1,20] is divided into n subintervals of length deltax, where rk is the right endpoint of the kth subinterval. which definite integral is equivalent to (as n approaches infinity) limit sigma n, k=1 2sqrt(1+rk) deltax? integrate 1 to 20 2x sqrt(1+x) dx integrate 1 to infinity 2x sqrt(1+x) dx integrate 1 to 20 2 sqrt(1+x) dx integrate 1 to infinity 2 sqrt(1+x) dx

integrate 1 to 20 2 sqrt(1+x) dx

the solution to dy/dx = sec x tan x with initial condition (pi/6, 1) is y = tan x + 1 sec x + 1 tan x - sqrt3/3 + 1 sec x - (2sqrt3)/3 + 1

sec x - (2sqrt3)/3 + 1

two students wrote antiderivative formulas for k sun (kx), where k > 1 student 1: integral of (k sin(kx))dx = k^2 cos (kx) + C student 2 integral of (k sin(kx))dx = -cos (kx) +C what student, if any, wrote a correct antiderivative formula student 1 only student 2 only both student 1 and 2 neither student 1 or 2

student 2 integral of (k sin(kx))dx let u = kx du = k * dx dx = du/k = integral of k* (1/k) sin u du = integral of sin u du = -cosu + c = -cosu (kx) + c = student 2's answer

Region R is bounded by the functions f(x) = log x and g(x) = 1/9(x^2-10x + 9). which statement describes the area of region R? the area is approximately 14.595 the area is approximately 15.091 the area is approximately 15.573 the area cannot be found as limit f(x) as x approaches 0 is -infinity

the area is approximately 15.091

Which statement describes limit as x approaches 0 of (sinx - 2x)/(4x^3) the limit is - infinity the limit is 0 the limit is -1/24 the limit DNE

the limit is - infinity