Warm-Up
Jan 17, 2016
Warm-Up
4-1: Antiderivatives & Indefinite Integrals
©2002 Roy L. Gover (www.mrgover.com)
Objectives:•Define the antiderivative (indefinite integral)•Learn basic antidifferentiation rules•Solve simple differential equations
Definition
If F(x) is an antiderivative of f(x) on an interval I, then F’ (x)=f(x) + c where c is a constant.
Example
Let be an antiderivative of f(x) then
3( ) 2F x x
2'( ) 6 ( )F x x f x c
The Problem...
Find the function given its derivative. The function is the antiderivative or indefinite integral.
Example
'( )f x xthe function whose derivative is . Confirm your answer.
x
If ,find f(x),
Try This2'( )f x x
the function whose derivative is . Confirm your answer.
2x
31( )
3f x x c
If ,find f(x),
Try This3'( )f x x
the function whose derivative is . Confirm your answer.
3x
41( )
4f x x c
If ,find f(x),
Try This5'( )f x x
the function whose derivative is . Confirm your answer.
5x
61( )
6f x x c
If ,find f(x),
Try This'( ) nf x x
the function whose derivative is .
nx
If ,find f(x),
11( )
1nf x x c
n
aka the power rule for integration.
Example
Find the antiderivative of
2x2.
Example
Find the antiderivative of cos x.
Try This
Find the antiderivative of sin x.
-cos x+c
Example
Find the general solution of the differential equation:
4dy
dx
The general solution to:
is: y=4x+c
because:
4dy
dx
[4 ]4
d x c
dx
Multiple solutions exist because:
Important Idea
[4 0]4
d x
dx
[4 6]4
d x
dx
[4 12]4
d x
dx
4 Solutions to :
0
10
20
30
40
1 2 3 4 5
c=0
c=6
c=12
c=18
4dy
dx y=4x+c
Definition
( )F f x dx c Integrand
Variable of Integration
Constant of Integration
Integral Sign
Example
Find the antiderivative:
2(2 3 4)x x dx
Power Rule for Integration1
1
nn x
x dx cn
1n
Special Case:
0 1 101
0 1 1
x xdx dx x dx x
Definition
Sum& Difference Rule for Integration
[ ( ) ( )] ( ) ( )f x g x dx f x dx g x dx The integral of the sum is the sum of the integrals…
Definition
The integral of the difference is the difference of the integrals
ExampleEvaluate the indefinite
integral:
5
1dx
x
Try ThisFind the antiderivative:
3 2( 2)x dxHint: re-write and use the power rule.
5
332
5x x c
Assignment
page 255 9-14 all,15-27 odd
Assignment
page 255 9-14 all,15-27 odd
Warm-Up Find the antiderivative:
3
( )x x
dxx
Hint: Re-write as 2 fractions.
Solution
3
3
xx c
ExampleEvaluate the indefinite integral:
2
cos
1 cos
xdx
x
Lesson CloseSee page 250 of your text for basic integration rules.
Since integration is the inverse of differentiation, you already know most of the rules.
Memor
iz
e
Intro to Differential Equations.
We need to do some work on Differential equations…
ExampleSometimes we want a particular antiderivative...
Find the equation for y given that y’=2x+3 and y passes through the point (2,1)
ExampleFind the equation for y given that y’=2x+3 and y passes through the point (2,1)
You try
Find the general solution of F'(x) = x3 + 2x and the particular solution passing through point (2, 6) Note: (2, 6) means F(2) = 6
Particular SolutionsFind the general solution of F'(x) = x3 + 2x and the particular solution passing through point (2, 6)F(x) = ∫x3 + 2x = x4 + x2 + C 4F(2) = 6 = 24 + 22 + C 4
general solution
F(x) = x4 + x2 - 2 4
particular solution
-2 = c
Solving a Problem
A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?
Assignment
page 255 #35-41odd, 55-61 odd, 63, 69.
QR code activity1) If you do not have a QR code scanner, partner with someone who does.2) Choose a random QR Code to scan, and solve the given problem.3) In the appropriate blank, record your answer and justification on the back of your worksheet.4) Use that answer to find the next problem. (e.g. The answer to #7 should be at the bottom of the QR code for #8)
Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?
s' = ∫s'' = ∫-32 feet/secposition = s velocity = s' acceleration = s''
s' = -32t + c
How can we find c?
Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?
position = s velocity = s' acceleration = s'' s' = -32t + c
initial velocity is 64 when time is zero 64 = -32(0) + c
64 = c
s' = -32t + 64
Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?
position = s velocity = s' acceleration = s'' s = ∫s' = ∫-32t + 64
s = -32t2 + 64t + c 2How do we find c this time?
Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?
position = s velocity = s' acceleration = s'' s = -32t2 + 64t + c
2initial height is 80 (time is 0)
80 = -32(0)2 + 64(0) + c 280 = c
Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?
position = s velocity = s' acceleration = s''
80 = -32(0)
2 + 64(0) + c
280 = cs = -16t2 +64t + 80
So when does the ball hit the ground?
Solving a ProblemA ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet.a) Find the position function giving the height s as a function of time t.b) When does the ball hit the ground?s = -16t2 +64t + 80
The position will be zero.
0 = -16t2 +64t + 800 = -16(t - 5)(t + 1) t = 5 or -1