1. Approximate
Integration
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2. Integration Techniques
 There are two situations in which it is impossible to find
the exact value of a definite integral.
 The first situation arises from the fact that, in order to
evaluate using the Fundamental Theorem of
Calculus (FTC), we need to know an anti-derivative of f.
 However, sometimes, it is difficult, or even impossible, to
find an anti-derivative.
 For example, it is impossible to evaluate the following
integrals exactly:
 The second situation arises when the function is
determined from a scientific experiment through
instrument readings or collected data.
 There may be no formula for the function :
21 1
3
0 1
1x
e dx x dx
( )
b
a
f x dx
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3. Approximate Integration
 We already know one method for approximate integration.
 Recall that the definite integral is defined as a limit of Riemann
sums.
 So, any Riemann sum could be used as an approximation to the
integral.
 If we divide [a, b] into n subintervals of equal length
∆x = (b – a)/n, we have:
where xi* is any point in the i th subinterval [xi -1, xi].
Ln Approximation
 If xi* is chosen to be the left endpoint of the
interval, then xi* = xi -1 and we have:
 The approximation Ln is called the left endpoint approximation.
1
( ) ( *)
nb
ia
i
f x dx f x x
1
1
( ) ( )
nb
n ia
i
f x dx L f x x
------- Equation 1
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4.  If f(x) ≥ 0, the integral represents
an area and Equation 1 represents
an approximation of this area by
the rectangles shown here.
Rn Approximation
 If we choose xi* to be the right endpoint, xi* = xi and we
have:
 The approximation Rn is called right endpoint approximation.
1
( ) ( )
nb
n ia
i
f x dx R f x x
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5.  The figure shows the midpoint approximation Mn.
 Mn appears to be better than either Ln or Rn.
Mn Approximation
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6. Midpoint Rule
 Let f be continuous on [a, b].
 The Midpoint Rule for approximating
is given by
x
y
a 1
x 2
x bnx…
)(xfy
b
a
dxxf )(
1 2
( )
[ ( ) ( ) ... ( )]
b
na
n
f x dx M
x f x f x f x
b a
x
n
1
1 12 ( ) midpoint of [ , ]i i i i ix x x x x
Where
and

7. .72135.3235.2235.12
333
EXAMPLE
SOLUTION
Approximate the following integral by the midpoint rule.
We have Δx = (b – a)/n = (4 – 1)/3 = 1. The
endpoints of the four subintervals begin at a = 1 and
are spaced 1 unit apart. The first midpoint is at a +
Δx/2 = 1.5. The midpoints are also spaced 1 unit
apart. According to the midpoint rule, the integral is
approximately equal to
3;32
4
1
3
ndxx
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8.  Let f be continuous on [a, b]. The Trapezoidal Rule for
approximating is given by
Trapezoidal Rule
b
a
dxxf )(
0 1 2
1
( )
( ) 2 ( ) 2 ( )
2
... 2 ( ) ( )
b
na
n n
f x dx T
x
f x f x f x
f x f x
where ∆x = (b – a)/n and xi = a + i ∆x
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9.  The reason for the name can be seen from the figure, which
illustrates the case f(x) ≥ 0.
 The area of the trapezoid that lies above the i th subinterval
is:
 If we add the areas of
all these trapezoids,
we get the right side of
the Trapezoidal Rule.
1
1
( ) ( )
[ ( ) ( )]
2 2
i i
i i
f x f x x
x f x f x
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10. .90
2
1
34233223222312
3333
EXAMPLE
SOLUTION
Approximate the following integral by the trapezoidal rule.
As in the last example, Δx = 1 and the endpoints of the
subintervals are a0 = 1, a1 = 2, a2 = 3, and a3 = 4. The
trapezoidal rule gives
3;32
4
1
3
ndxx
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11. Simpson’s Rule
 Rather than using straight lines to approximate the
curve, Simpson’s Rule uses parabolas.
0 1 2 3
2 1
( )
[ ( ) 4 ( ) 2 ( ) 4 ( )
3
... 2 ( ) 4 ( ) ( )]
b
na
n n n
f x dx S
x
f x f x f x f x
f x f x f x
Where n is even
and ∆x = (b – a)/n.
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12. EXAMPLE
SOLUTION
Putting f(x) = 1/x, n = 10, and ∆x = 0.1 in Simpson’s Rule,
we obtain:
Use Simpson’s Rule with n = 10 to approximate
2
1
(1/ )x dx
2
101
1
[ (1) 4 (1.1) 2 (1.2) 4 (1.3)
3
... 2 (1.8) 4 (1.9) (2)]
1 4 2 4 2 4 2 4
0.1 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7
2 4 13
1.8 1.9 2
0.693150
x
dx S f f f f
x
f f f
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13. The End
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