chapter 6 (applications of integrals)

Question 1

find positionat the end of a given interval given the velocity equation

Answer 1

it is the integral over the given bounds of the velocity equations

Question 2

given the initial position and an equation for the velocityfind the final positon

Answer 2

integrate the velocity equation

find c by plugging in the initial position 1

Question 3

find the velocity equation given the speed at a certain time and an equation for acceleration

Answer 3

the equation is found by integrating the acceleration equation and finding c with the given velocity and time

Question 4

find the net change of Q that changes over time

Answer 4

this is the integral from a to b of

Question 5

finding the area in between curves on the bounds of a to b

Answer 5

the integral from a to b of the greater function minus the smaller function int(a to b) (fx-gx)dx

Question 6

how do you determine the greater function

Answer 6

whichever one is higher up

Question 7

what if there is an absolute value as one of the curves

Answer 7

you must find the area in 2 separate parts the negative will end up canceling out on the left side of the curve.

Question 8

how to find the area in between curves with respect to y

Answer 8

switch the functions to be y dependent
find the greater function relative to the y axis
integrate(g(y)-l(y))

Question 9

general slicing method

Answer 9

a way to find area using general slices

Question 10

how to use the general slicing method

Answer 10

1. find the function for the area of a cross section
2. find the area function in terms of the determanant given function
3. integrate on the bounds of the function.

int(a to b) of A(r(x))dx

Question 11

disk method

Answer 11

used to find volume of a shape by rotating it about an axis

Question 12

disk method how to

Answer 12

1. find the function that determines the radius relative to the axis
2. use the equation int(atob) (pi*(radius equation)^2)

Question 13

washer method

Answer 13

a section is rotated about an axis however there is a gap in between them

Question 14

washer method how to

Answer 14

1. find the radius that determines the outer edge of the washer. (fx)
2. find the radius that determines the inner side of the area (gx)
3. find the bounds (the width of the washer)
4. use dis int(atob) pi*(f(x)^2-g(x)^2)

Question 15

washer method on the y axis.

Answer 15

do it the same way as on the x axis but with a function relative to y

Question 16

shell method

Answer 16

rotating about the x axis wit a function that is relative to y. it uses the theory of circumference.

Question 17

shell method how to

Answer 17

1. find the axis and the region
2. find the radius… this can be just x or y
3. find the equation for the height. this will be parallel sections to the axis of rotation
4. use this int(atob) (2(pi)(radius)*(function for the height)

Question 18

disk method equation

Answer 18

int(atob) pi*(radius equation)^2

Question 19

washer method equation

Answer 19

int(atob) pi*(big radius^2-small radius^2)

Question 20

shell method equation

Answer 20

int(atob)2piradius*(height fucntion)

Question 21

length of curve equation

Answer 21

int(atob) sqrt(1+f’(x)^2)

Question 22

surface area defined by x rotated about x

Answer 22

int(atob) 2pi(fx)*(sqrt(1+f’(x)^2))

Question 23

surface area defined by x rotated about y axis

Answer 23

int(atob) 2pi(x)*(sqrt(1+f’(x)^2))

Question 24

surface area defined by y rotated about y axis

Answer 24

int(atob) 2pi(gy)*(sqrt(1+g’(y)^2))

Question 25

surface are defined by y about the x axis

Answer 25

int(atob) 2pi(y)*(sqrt(1+g’(y)^2))

Question 26

pumping equation

Answer 26

W=int(liquid start height to liquid end height) of (densitygravityareadistance to move)