Indefinite Integrals, ApplicationsSection 6.1b

The set of all antiderivatives of a function isthe indefinite integral of with respect to and isdenoted by

f x dxIntegral Sign

IntegrandVariable ofIntegration

Also, recall that a function is an antiderivativeof if F x f x

Definition: Indefinite Integral f x

f x

F x f x

Definition: Indefinite Integral

Then all antiderivatives of a function vary by constants:

f x dx F x C What keeps this integral from being “definite”???

The constant C is the constant of integration andis an arbitrary constant.

When we find we have integrated or evaluated the integral…

F x C f

Integral FormulasIndefinite Integral

1

1

nn xx dx C

n

Reversed Derivative Formula

1. (a)

1n

1

1

nnd xx

dx n

1n

lndx

x Cx

(b)1

lnd

xdx x

kxkx ee dx C

k 2.

kxkxd ee

dx k

Integral FormulasIndefinite Integral Reversed Derivative Formula

cossin

kxkx dx C

k 3.

cossin

d kxkx

dx k

sincos

kxkx dx C

k 4.

sincos

d kxkx

dx k

2sec tanx dx x C 5. 2tan secd

x xdx

Integral FormulasIndefinite Integral Reversed Derivative Formula

2csc cotx dx x C 6. 2cot cscd

x xdx

sec tan secx x dx x C 7. sec sec tand

x x xdx

csc cot cscx x dx x C 8.

csc csc cotd

x x xdx

Using Integral FormulasEvaluate:

5x dx6

6

xC

1dxx 1 2x dx 1 22x C 2 x C

3xe dx3

3

xeC

31

3xe C

cos2

xdx

sin 1 2

1 2

xC 2sin

2

xC

Properties of Indefinite Integrals

kf x dx k f x dx Let k be a real number.

1. Constant Multiple Rule:

f x dx f x dx If k = –1, then:

f x g x dx f x dx g x dx 2. Sum and Difference Rule:

Integrating Term by Term

2 2 5x x dx Evaluate

2 2 5x dx xdx dx 3

21 2 35

3

xC x C x C

But we can simply combine all of these constants!!!

3

2 22 5 53

xx x dx x x C

Do Now – p.314, #55How long did it take the hammer and feather to fall 4 ft on themoon? Solve the following initial value problem for s as afunction of t. Then find the value of t that makes s equal to 0.

Differential equation:

22

25.2ft sec

d s

dt

Initial conditions: and when0ds

dt 4s 0t

2

25.2

d sdt dt

dt

15.2ds

t Cdt

10 5.2 0 C 1 0C

Velocity: 5.2ds

tdt

Do Now – p.314, #55How long did it take the hammer and feather to fall 4 ft on themoon? Solve the following initial value problem for s as afunction of t. Then find the value of t that makes s equal to 0.

Differential equation:

22

25.2ft sec

d s

dt

Initial conditions: and when0ds

dt 4s 0t

5.2dsdt tdt

dt

222.6s t C

2 24 2.6 0 C 2 4C

Position: 22.6 4s t t

Do Now – p.314, #55How long did it take the hammer and feather to fall 4 ft on themoon? Solve the following initial value problem for s as afunction of t. Then find the value of t that makes s equal to 0.

Differential equation:

22

25.2ft sec

d s

dt

Initial conditions: and when0ds

dt 4s 0t

22.6 4s t t Solving , we have 0s t 2 4

2.6t

Take the positive solution… 1.240t They took about 1.240 seconds to fall

More Application ProblemsA right circular cylindrical tank with radius 5 ft and height 16 ftthat was initially full of water is being drained at the rate of0.5 x ft /min (x = water’s depth). Find a formula for the depthand the amount of water in the tank at any time t. How long willit take the tank to empty?

3

x

2V r h 25V x 25 x

Diff Eq: 25dV dx

dt dt

0.5 25dx

xdt

More Application ProblemsA right circular cylindrical tank with radius 5 ft and height 16 ftthat was initially full of water is being drained at the rate of0.5 x ft /min (x = water’s depth). Find a formula for the depthand the amount of water in the tank at any time t. How long willit take the tank to empty?

3

50

dx x

dt

Initial Condition:

0 16x

1 2 1

50

dxx

dt

Solve Analytically:

1 2 1

50

dxx dt dt

dt

1 2 1

50x dx dt

Diff Eq: 0.5 25dx

xdt

More Application ProblemsA right circular cylindrical tank with radius 5 ft and height 16 ftthat was initially full of water is being drained at the rate of0.5 x ft /min (x = water’s depth). Find a formula for the depthand the amount of water in the tank at any time t. How long willit take the tank to empty?

3

1 2 12 16 0

50C

Solve Analytically:

8C

1 2 12

50x t C

1 2 1

50x dx dt

Initial Condition:

1 2 12 8

50x t

2

4100

tx

More Application ProblemsA right circular cylindrical tank with radius 5 ft and height 16 ftthat was initially full of water is being drained at the rate of0.5 x ft /min (x = water’s depth). Find a formula for the depthand the amount of water in the tank at any time t. How long willit take the tank to empty?

3

400t

Equation for volume: 2

25 25 4100

tV x

At what t is V = 0? minutes

(The tank will be empty in about 21 hours)

More Application ProblemsYou are driving along a highway at a steady 60 mph (88 ft/sec)when you see an accident ahead and slam on the brakes. Whatconstant deceleration is required to stop your car in 242 feet?

First, solve the following initial value problem:2

2

d sk

dtDifferential Equation: (k constant)

88ds

dtInitial Conditions: 0s and 0t when

88ds

ktdt

Velocity:

2

882

kts t Solution:

More Application ProblemsYou are driving along a highway at a steady 60 mph (88 ft/sec)when you see an accident ahead and slam on the brakes. Whatconstant deceleration is required to stop your car in 242 feet?

Next, find the value of t that makes ds/dt = 0:

88ds

ktdt

Velocity:

2

882

kts t Solution:

88 0kt 88

tk

More Application ProblemsYou are driving along a highway at a steady 60 mph (88 ft/sec)when you see an accident ahead and slam on the brakes. Whatconstant deceleration is required to stop your car in 242 feet?

88ds

ktdt

Velocity:

2

882

kts t Solution:

Finally, find the value of k that makes s = 242 for the previouslyfound value of t :

88242s

k

288 88

88 2422

k

k k

3872242

k

216ft seck You would need to decelerate at thisconstant rate in order to stop in 242 feet!!!