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Chapter 4
Integration
Indefinite Integral
or
Antiderivative
n 1
x
x
dx


c
,
n


1

n 1
n
4
x


3
 2 x  dx  x  x  c
4
2
Find the Particular Solution
or
Solve the Differential Equation
dy
 4 x  1, at (1,5)
dx
2
y  2x  x  4
1. Change into a differential equation.
2. Integrate both sides of the equation.
3. Find c by plugging in the coordinate.
4. Replace c and write the particular
solution.
st
1
Fundamental
Theorem of Calculus
Definite Integral
or
Area Under the Curve
on the interval [a,b]

b
a
f ( x)dx  F (b)  F (a )
4
x

2
dx

(3)



1
3
2
 2(3)    (1)  2(1)   4
2
Area below the x-axis is
NEGATIVE
Approximate the
Area Under a Curve
Using a Left-Sided Sum
f ( x)   x  6x on 1, 4 with 3subintervals
2
4 1
A
 f (1)  f (2)  f (3)
3
Approximate the
Area Under a Curve
Using a Right-Sided Sum
f ( x)   x  6x on 1, 4 with 3subintervals
2
4 1
A
 f (2)  f (3)  f (4)
3
Approximate the
Area Under a Curve
Using a Midpoint Sum
f ( x)   x  6x on 1, 4 with 3subintervals
2
4 1
A
 f (1.5)  f (2.5)  f (3.5)
3
Midpoint Sum
Approximate the
Area Under a Curve
Using a Trapezoid Sum
f ( x)   x  6x on 1, 4 with 3subintervals
2
4 1 1
A
 f (1)  2 f (2)  2 f (3)  f (4)
3 2
Mean Value Theorem (MVT)
or
Average Value
Mean Value Theorem
b
fc
()
(
ba
)
fx
()
d
x

a
Average Value
1 b
fc
() f()
xd
x
a
ba

Find the x value
where you get the
Average Value
f ( x)  2 x  1,  2, 4
4
1
2x 1 
2
x

1
dx



42 2
1. Find the Average Value.
2. Set the original function equal
to the Average Value.
3. Solve for x.
nd
2
Fundamental
Theorem of Calculus
d  u

f
(
t
)
dt

f
(
u
)
u
'



a


du
2
d 

2
3
2
8
t
dt

4
x
12
x

192
x







1
dx 

4 x3
a

a
f (x)dx
0
a

b
f (x)dx
b
 f (x)dx
a
f
(
x
)

kd
x



a
b
b
b
a
a
f
()
xd
x

k
d
x


Integrate an Even Function

a
a
a
f ( x) dx  2  f ( x) dx
0
Integrate an Odd Function

a
a
f ( x) dx  0
U-Substitution
or
Change of Variables

f ( g ( x))g '( x) dx   f (u ) du  F (u )  C
2
x
x

3
dx

(
u
)
du




2
3
3
u  x  3 and du  2 x dx
2
Find Definite Integral Using
U-Substitution
or
Change of Variables


2
0
b
a
f ( g ( x))g '( x) dx  
2 x  x  3 dx  
2
3
g (2)
g (0)
g (b )
g (a)
f (u ) du
1
(u ) du   (u ) du
3
3
u  x  3 and du  2 x dx
2
3