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![Page 1: 1 Chapter 7 Integral Calculus The basic concepts of differential calculus were covered in the preceding chapter. This chapter will be devoted to integral.](https://reader034.fdocuments.us/reader034/viewer/2022052701/56649ed05503460f94bde10c/html5/thumbnails/1.jpg)
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Chapter 7Integral Calculus
The basic concepts of differential calculus were covered in the preceding chapter. This chapter will be devoted to integral calculus, which is the other broad area of calculus. The next chapter will be devoted to how both differential and integral calculus manipulations can be performed with MATLAB.
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Anti-Derivatives
An anti-derivative of a function f(x) is a new function F(x) such that
( )( )
dF xf x
dx
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Indefinite and Definite Integrals
( )f x dx2
1
( )x
xf x dx
Indefinite
Definite
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Definite Integral as Area Under the Curve
Approximate Area kk
y x
1y 2y3y
4y
Ky
a b x
y
5y
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Exact Area as Definite Integral
limb
ka x dxk
ydx y x
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Definite Integral with Variable Upper Limit
x
aydx
( )x
ay u du
More “proper” form with “dummy” variable
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Area Under a Straight-Line Segment
1x 2x x
1y
2y( )y f x
L
2
12 1 2 1
1
2
x
xydx y y x x
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Example 7-1. Determine12
0ydx
20 20
-20
-10
24 6
89 12
x
( )y f x
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Example 7-1. Continuation.
12
040 0 20 20 0 30 10ydx
20 20
-20
-10
24 6
89 12
x
( )y f x
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Example 7-2. Determine0
xydx
20 20
-20
-10
24 6
89 12
x
( )y f x
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Guidelines
1. If y is a non-zero constant, integral is either increasing or decreasing linearly.
2. If segment is triangular, integral is increasing or decreasing as a parabola.
3. If y=0, integral remains at previous level.4. Integral moves up or down from previous
level; i.e., no sudden jumps.5. Beginning and end points are good
reference levels.
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20 20
-20
-10
24 6
89 12
x
( )y f x(a)
40
20
40
10
x2 4 6 8 9 12
0 0( )
x xydx f x dx (b)
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Tabulation of Integrals
( ) ( )F x f x dx( )
b
aI f x dx
( ) ( ) ( )b
aI F x F b F a
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Table 7-1. Common Integrals.( )f x ( ) ( )F x f x d x I n t e g r a l N u m b e r
( )a f x ( )a F x I - 1
( ) ( )u x v x ( ) ( )u x d x v x d x I - 2
a a x I - 3
1nx n 1
1
nx
n
I - 4
a xe a xe
a
I - 5
1
x l n x I - 6
s i n a x1
c o s a xa
I - 7
c o s a x1
s i n a xa
I - 8
2s i n a x1 1
s i n 22 4x a x
a I - 9
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Table 7-1. Continuation.
2c o s a x1 1
s in 22 4x a x
a I -1 0
s inx a x 2
1s in c o s
xa x a x
a a I -1 1
c o sx a x 2
1c o s s in
xa x a x
a a I -1 2
s in c o sa x a x 21s in
2a x
aI -1 3
s in c o sa x b x2 2fo r a b
c o s ( ) c o s ( )
2 ( ) 2 ( )
a b x a b x
a b a b
I -1 4
a xx e 21
a xea x
a I -1 5
ln x ln 1x x I -1 6
2
1
a x b11
ta na
xba b
I -1 7
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In Examples 7-3 through 7-5 that follow, determine the following integral in each case:
z ydx
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Example 7-3
412 xy e4
4
4
12 124
3
xx
x
ez e dx C
e C
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Example 7-4
12 sin 2y x x
2
12 sin 2
112 sin 2 cos 2
(2) 2
3sin 2 6 cos 2
z x xdx
xx x C
x x x C
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Example 7-5
2 36y x
x
2
2
3
3
36
36
63ln
3
2 3ln
z x dxx
x dx dxx
xx C
x x C
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In Examples 7-6 and 7-7 that follow, determine the definite integral in each case as defined below.
b
aI ydx
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Example 7-6
0sinI xdx
00sin cos
cos cos 0
( 1) ( 1) 2
I xdx x
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Example 7-7
1 2
08 xI xe dx
1 2
0
21
2 0
2 0
2
8
8 2 1(2)
2 2(1) 1 2 0 1
6 2 1.188
x
x
I xe dx
ex
e e
e
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Displacement, Velocity, and Acceleration
2 2( ) acceleration in meters/second (m/s )
( ) velocity in meters/second (m/s)
( ) displacement in meters m
a a t
v v t
y y t
( )dv
a tdt
( )dv
dv dt a t dtdt
( )dv a t dt dv v
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Displacement, Velocity, and AccelerationContinuation
1( )v a t dt C ( )
dyv t
dt ( )
dydy dt v t dt
dt
2( )y v t dt C
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Alternate Formulation in Terms of Definite Integrals
0( ) ( ) (0)
tv t a t dt v
0( ) ( ) (0)
ty t v t dt y
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Example 7-8. An object experiences acceleration as given by
2( ) 20 ta t e
Determine the velocity and displacement.
(0) 0
(0) 0
v
y
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Example 7-8. Continuation.
2( ) 20 tdva t e
dt
2 2 21 1
20( ) 20 10
2t t tv t e dt e C e C
0
1 1(0) 10 10 0v e C C
1 10C 2( ) 10 10 tv t e
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Example 7-8. Continuation.
2
2 22 2
( ) 10 10
1010 10 5
2
t
t t
y t e dt
t e C t e C
02 2(0) 0 5 5 0y e C C
2 5C
( ) 10 5 5ty t t e
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Example 7-9. Rework previous example using definite integral forms.
2 2
0 00
2 0 2
20( ) ( ) (0) 20 0
2
10 10 10 10
tt t t t
t t
v t a t dt v e dt e
e e e
2 2
0 0 0
2 0 2
( ) ( ) (0) (10 10 ) 10 5
10 5 0 5 10 5 5
tt t t t
t t
y t v t dt y e dt t e
t e e t e
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Example 7-10. Plot the three functions of the preceding examples.
2( ) 20 ta t e2( ) 10 10 tv t e
2( ) 10 5 5ty t t e
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Example 7-10. Continuation.
>> t = 0:0.02:2;>> a = 20*exp(-2*t);>> v = 10 -10*exp(-2*t);>> y = 10*t + 5*exp(-2*t) - 5;>> plot(t, a, t, v, t, y)
The plots are shown on the next slide.
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