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Review 5.1-5.3
Distance Traveled
Area Under a curve
Antiderivatives
Calculus
time
velocity
After 4 seconds, the object has gone 12 feet.
Consider an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance:
If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
ft3 4 sec 12 ft
sec
3t d
If the velocity is not constant,we might guess that the distance traveled is still equalto the area under the curve.
(The units work out.)
211
8V t Example:
We could estimate the area under the curve by drawing rectangles touching at their left corners.
This is called the Left-hand Rectangular Approximation Method (LRAM).
1 11
8
11
2
12
8t v
10
1 11
8
2 11
2
3 12
8Approximate area: 1 1 1 3
1 1 1 2 5 5.758 2 8 4
We could also use a Right-hand Rectangular Approximation Method (RRAM).
11
8
11
2
12
8
Approximate area: 1 1 1 31 1 2 3 7 7.75
8 2 8 4
3
211
8V t
Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).
1.031251.28125
1.78125
Approximate area:6.625
2.53125
t v
1.031250.5
1.5 1.28125
2.5 1.78125
3.5 2.53125
In this example there are four subintervals.As the number of subintervals increases, so does the accuracy.
211
8V t
211
8V t
Approximate area:6.65624
t v
1.007810.25
0.75 1.07031
1.25 1.19531
1.382811.75
2.25
2.75
3.25
3.75
1.63281
1.94531
2.32031
2.75781
13.31248 0.5 6.65624
width of subinterval
With 8 subintervals:
The exact answer for thisproblem is .6.6
Circumscribed rectangles are all above the curve:
Inscribed rectangles are all below the curve:
When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
211
8V t
subinterval
The width of a rectangle is called a subinterval.
Subintervals do not all have to be the same size.
b
af x dx
IntegrationSymbol
lower limit of integration
upper limit of integration
integrandvariable of integration
(dummy variable)
It is called a dummy variable because the answer does not depend on the variable chosen.
b
af x dx
We have the notation for integration, but we still need to learn how to evaluate the integral.
time
velocity
After 4 seconds, the object has gone 12 feet.
In section 5.1, we considered an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance: 3t d
If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
ft3 4 sec 12 ft
sec
If the velocity varies:
11
2v t
Distance:21
4s t t
(C=0 since s=0 at t=0)
After 4 seconds:1
16 44
s
8s
1Area 1 3 4 8
2
The distance is still equal to the area under the curve!
Notice that the area is a trapezoid.
Fill in the blanks:
a)Integrating velocity gives ____________________
b)Integrating the absolute value of velocity gives ______________
c) New Position = _____________ + _______________
Displacement
Total Distance
Initial Disp. Total Distance
Definition: A function F is called an antiderivative of f on an interval I if F’(x) = f(x) for all x in I.
Example: Let f(x)=x3. If F(x) =1/4 * x4 then F’(x) = f(x)
Theorem: If F is an antiderivative of f on an interval I, then the most generalantiderivative of f on I is
F(x) + Cwhere C is an arbitrary constant.
Antiderivatives
F is an antiderivative of f
Table of particular antiderivatives
Function Antiderivative Function Antiderivative
xr, r -1 xr+1/(r+1) 1/x ln |x|
sin x - cos x ex ex
cos x sin x ax ax / ln a
sec2 x tan x tan-1 x
sin-1 x 2
1
1 x
2
1
1 x
These rules give particular antiderivatives of the listed functions. A general antiderivative can be obtained by adding a constant .
Integrate
Rewriting before integrating
Find the general solution of the equation F’(x) = and
find the particular solution given the point F(1) = 0.
Now plug in (1,0) and solve for C.
0 = -1 + C
C = 1
Final answer.
Exploration
Finding Antiderivatives For each of the following derivatives, describe the original
function .F
(a) '( ) 2F x x
(b) '( )F x x
2(c) '( )F x x
2
1(d) '( )F x
x
3
1(e) '( )F x
x
(f) '( ) cosF x x
2( )F x x
2( ) / 2F x x
3( ) / 3F x x
( ) 1/F x x
2( ) 1/ 2F x x
( ) sinF x x
2( ) 3f x x
Find the function that is the antiderivative of .F f
3( ) + C becauseF x x3
2[ ]3
d x Cx
dx
2
Here are some members of the
family of antiderivatives of ( ) 3 :f x x
31( ) 5F x x 3
2 ( ) 36F x x ...etc
33( )
3
xF x C
2[ ] 2xD x x
The family of all antiderivatives of ( ) 2 is
represented by
f x x
2( )F x x C
Find the general solution of the differential equation
' 2y 2y x C
:Notation
( )y f x dx ( )F x C
Integrand
Variable of
Integration
Constant of
Integration
is the antiderivative (indefinite integral) of ( )
with respect to .
y f x
x
3xdx 3 xdx 2
32
xC
23
2x C
RewriteIntegrate
Simplify
3
1dx
x 3x dx 2
1
2C
x
2
2
xC
xdx 1/ 2x dx 3 / 2
3/ 2
xC
3 / 22
3x C
2sin xdx 2 sin xdx 2 ( cos )x C
2cos x C
( 2)x dx 2
1 222
xC x C
2xdx dx
2
22
xx C
4 2(3 5 )x x x dx May we skip step 2?????
5 3 23 5 1
5 3 2x x x C
5 3 2
3 55 3 2
x x xC
1xdx
x
1/ 2 1/ 2( )x x dx 3 / 2 1/ 2
3/ 2 1/ 2
x xC 3 / 2 1/ 22
23
x x C
Never integrate the numerator and denominator seperately!
Rewrite the Integrand
2 5 3 2x xdx
x
5/2 3/2 1/212 2220
5 3x x x C
2
sin
cos
xdx
x
1 sin
cos cos
xdx
x x
sec tanx xdx sec x C
????
Find the general solution of
2
1'( ) , 0f x F x x
x
2( )F x x dx1
1
xC
1
Cx
Find the particular solution that
satisfies the initial condition
(1) 0F 1
(1)1
F C 0
1C
1( ) 1, 0F x x
x
1F x C
x
Solve the differential equation given the initial condition.
3 1, 2 3f x x f
231
2f x x x
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 as a function of the time .s t
0 initial timet
Given Initial Conditions:Acceleration due to gravity: 32 feet per second per second
''( ) 32s t
'(0) 64s (0) &80s
80 ft
(a) Find the position function giving the height as a function of the time .s t
0 initial timet Given Conditions:Acceleration due to gravity: 32 feet per second per second
''( ) 32s t
'(0) 64s (0) &80s
'( )s t ''( ) 32s t dt dt 132t C
164 32(0) C 1 64C
'( ) 32 64s t t
(a) Find the position function giving the height as a function of the time .s t
( ) '( ) ( 32 64)s t s t dt t dt 2
232 642
tt C
2216 64t t C Using (0) 80:s
2
280 16 0 64 0 C 2 80C 2( ) 16 64 80s t t t
(b) When does the ball hit the ground?
2( ) 64 816 0s t t t 0216( 4 5) 0t t
16( 5)( 1) 0t t
5 secondst
80 ft
5t
Basic Integration Rules
These two equations allow you to obtain integration formulas directly from differentiation formulas, as shown in the following summary.
Basic Integration Rulescon
t’d
Before you begin the exercise set, be sure you realize
that one of the most important steps in integration is
rewriting the integrand
in a form that fits the basic integration rules.
Practice Exercises
Original Rewrite Integrate Simplify
2dx
x
22 1t dt
3
2
3xdx
x
3 4x x dx
Computing AntiderivativesProblem
2
cos 3Let f . Find antiderivatives of the function f.
12
xx
x
Solution
2
An antiderivative of the function cos is sin
1and an antiderivative of is arctan .
1
x x
xx sin
Hence an antiderivative of the given function is 3arctan .2
xx
Analyzing the motion of an object using antiderivatives
A particle is moving with the given data. Find the position of the particle.a(t) = 10 + 3t -3t2, v(0) = 2, s(0) = 5
v(t) is the antiderivative of a(t): v’(t) = a(t) = 10 + 3t -3t2
Antidifferentiation gives v(t) = 10t + 1.5t2 – t3 + Cv(0) = 2 implies that C=2; thus, v(t) = 10t + 1.5t2 – t3 + 2
s(t) is the antiderivative of v(t): s’(t) = v(t) = 10t + 1.5t2 – t3 + 2Antidifferentiation gives s(t) = 5t2 + 0.5t3 – 0.25t4 + 2t + Ds(0) = 5 implies that D=5; thus, s(t) = 5t2 + 0.5t3 – 0.25t4 + 2t + 5
Problem
Solution
Some indefinite integrals
1
xx
2 2
1 + C , if n -1. ln(| |)
1 x
a, a
ln(a)
sin( ) cos( ) , cos( ) sin( ) ,
sec ( ) tan( ) , csc ( )
nn
x x
xx dx dx x C
n
e dx e C dx C
x dx x C x dx x C
x dx x C x dx
1 12 2
cot( ) ,
sec( ) tan( ) sec( ) , csc( ) cot( ) csc( ) ,
1 1tan ( ) , sin ( ) ,
1 1
x C
x x dx x C x x dx x C
dx x C dx x Cx x
