Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
Approximate Integration
MIDPOINT RULE:
Zb
f (x)dx ≈ Mn = ∆x[f (x1 ) + f (x2 ) + . . . + f (xn )]
a
where
∆x =
b−a
n
1
and xi = (xi−1 + xi ) = midpoint of [xi−1 , xi ].
2
TRAPEZOIDAL RULE:
Zb
a
f (x)dx ≈ Tn =
∆x
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + . . . + 2f (xn−1 ) + f (xn )]
2
where
∆x =
and xi = a + i∆x.
1
b−a
n
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
EXAMPLE: Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n = 5 to approximate
the integral
Z2
1
dx
x
1
Solution:
(a) With n = 5, a = 1, and b = 2, we have ∆x = (2 − 1)/5 = 0.2, and so the Trapezoidal Rule
gives
Z2
∆x
1
dx ≈ T5 =
[f (1) + 2f (1.2) + 2f (1.4) + 2f (1.6) + 2f (1.8) + f (2)]
x
2
1
0.2 1
2
2
2
2
1
≈ 0.695635
=
+
+
+
+
+
2 1 1.2 1.4 1.6 1.8 2
(b) The midpoints of the five subintervals are 1.1, 1.3, 1.5, 1.7, and 1.9, so the Midpoint Rule
gives
Z2
1
dx ≈ M5 = ∆x[f (1.1) + f (1.3) + f (1.5) + f (1.7) + f (1.9)]
x
1
1 1
1
1
1
1
≈ 0.691908
=
+
+
+
+
5 1.1 1.3 1.5 1.7 1.9
REMARK: Note that
Z2
1
2
1
dx = ln x = ln 2 ≈ 0.693147
1
x
therefore the errors in the Trapezoidal and Midpoint Rule approximations for n = 5 are
ET ≈ −0.002488 and EM ≈ 0.001239
We see that the size of the error in the Midpoint Rule is about half the size of the error in
the Trapezoidal Rule.
EXAMPLE: Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n = 10 to approximate the integral
Z1
2
ex dx
0
2
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
EXAMPLE: Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n = 10 to approximate the integral
Z1
2
ex dx
0
Solution:
(a) With n = 10, a = 0, and b = 1, we have ∆x = (1 − 0)/10 = 0.1, and so the Trapezoidal
Rule gives
Z1
∆x
[f (0) + 2f (0.1) + 2f (0.2) + . . . + 2f (0.8) + 2f (0.9) + f (1)]
2
0
0.1 02
2
2
2
2
2
e + 2e0.1 + 2e0.2 + . . . + 2e0.8 + 2e0.9 + e1 ≈ 1.467174693
=
2
2
ex dx ≈ T10 =
(b) The midpoints of the ten subintervals are 0.05, 0.15, 0.25, . . . , 0.85, 0.95, so the Midpoint
Rule gives
Z1
0
=
2
ex dx ≈ M10 = ∆x[f (0.05) + f (0.15) + f (0.25) + . . . + f (0.85) + f (0.95)]
1 0.052
2
2
2
2
e
+ e0.15 + e0.25 + . . . + e0.85 + e0.95 ≈ 1.460393091
10
REMARK: One can compute that
Z1
2
ex dx ≈ 1.462651746
0
therefore the errors in the Trapezoidal and Midpoint Rule approximations for n = 10 are
ET ≈ −0.004522947 and EM ≈ 0.002258655
We see that the size of the error in the Midpoint Rule is about half the size of the error in
the Trapezoidal Rule.
EXAMPLE: Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n = 10 to approximate the integral
Z1 √
1 + x3 dx
0
3
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
EXAMPLE: Use (a) the Trapezoidal Rule and (b) the Midpoint Rule with n = 10 to approximate the integral
Z1 √
1 + x3 dx
0
Solution:
(a) With n = 10, a = 0, and b = 1, we have ∆x = (1 − 0)/10 = 0.1, and so the Trapezoidal
Rule gives
Z1 √
∆x
[f (0) + 2f (0.1) + 2f (0.2) + . . . + 2f (0.8) + 2f (0.9) + f (1)]
2
0
√
√
√
√
√
0.1 √
1 + 03 + 2 1 + 0.13 + 2 1 + 0.23 + . . . + 2 1 + 0.83 + 2 1 + 0.93 + 1 + 13
=
2
1 + x3 dx ≈ T10 =
≈ 1.112332391
(b) The midpoints of the ten subintervals are 0.05, 0.15, 0.25, . . . , 0.85, 0.95 so the Midpoint
Rule gives
Z1 √
0
=
1 + x3 dx ≈ M10 = ∆x[f (0.05) + f (0.15) + f (0.25) + . . . + f (0.85) + f (0.95)]
√
√
√
√
1 √
1 + 0.053 + 1 + 0.153 + 1 + 0.253 + . . . + 1 + 0.853 + 1 + 0.953
10
≈ 1.111005559
REMARK: One can compute that
Z1 √
1 + x3 dx ≈ 1.111447979
0
therefore the errors in the Trapezoidal and Midpoint Rule approximations for n = 10 are
ET ≈ −0.000884412 and EM ≈ 0.000442420
We see that the size of the error in the Midpoint Rule is about half the size of the error in
the Trapezoidal Rule.
ERROR BOUNDS: Suppose |f ′′ (x)| ≤ K for a ≤ x ≤ b. If ET and EM are the errors in the
Trapezoidal and Midpoint Rules, then
|ET | ≤
K(b − a)3
12n2
and |EM | ≤
K(b − a)3
24n2
EXAMPLE: Give upper bounds for the errors ET and EM involved in the approximation of
Z2
1
dx with n = 5.
x
1
4
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
EXAMPLE: Give upper bounds for the errors ET and EM involved in the approximation of
Z2
1
dx with n = 5.
x
1
Solution: Note that
f ′ (x) = −
Since
1
x2
and f ′′ (x) =
2
x3
2
is a decreasing function on [1, 2], we have
x3
2
2
′′
|f (x)| = 3 ≤ 3 = 2
x
1
Therefore, taking K = 2, a = 1, b = 2, and n = 5 in the above error estimates, we obtain
|ET | ≤
2(2 − 1)3
1
K(b − a)3
=
=
≈ 0.0067
2
2
12n
12(5)
150
|EM | ≤
K(b − a)3
2(2 − 1)3
1
=
=
≈ 0.0033
2
2
24n
24(5)
300
and
REMARK: Note that these error estimates are bigger than the actual errors 0.002488 and
0.001239.
EXAMPLE: Give upper bounds for the errors ET and EM involved in the approximation of
Z1
2
ex dx with n = 10.
0
Solution: Note that
f ′ (x) = 2xex
2
2
2
and f ′′ (x) = 2ex + 4x2 ex
2
2
Since 2ex + 4x2 ex is an increasing function on [0, 1], we have
2
2
2
2
|f ′′ (x)| = 2ex + 4x2 ex ≤ 2e1 + 4(1)2 e1 = 6e
Therefore, taking K = 6e, a = 0, b = 1, and n = 10 in the above error estimates, we obtain
|ET | ≤
K(b − a)3
6e(1 − 0)3
e
=
=
≈ 0.01359140914
12n2
12(10)2
200
|EM | ≤
K(b − a)3
6e(1 − 0)3
e
=
=
≈ 0.006795704570
24n2
24(10)2
400
and
REMARK: Note that these error estimates are bigger than the actual errors 0.004522947 and
0.002258655.
EXAMPLE: Give upper bounds for the errors ET and EM involved in the approximation of
Z1 √
1 + x3 dx with n = 10.
0
5
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
EXAMPLE: Give upper bounds for the errors ET and EM involved in the approximation of
Z1 √
1 + x3 dx with n = 10.
0
Solution: Note that (see the Appendix)
3x2
f ′ (x) = √
2 1 + x3
and f ′′ (x) =
3x(x3 + 4)
4(1 + x3 )3/2
We now find increasing/decreasing intervals of f ′′ (x). Here is the graph of f ′′ (x):
To find a point of a local maximum, we note that (see the Appendix)
f ′′′ (x) = −
3(x6 + 20x3 − 8)
8(1 + x3 )5/2
One can check that f ′′′ (x) = 0 on [0, 1] at x ≈ 0.7320508076 which is a root of x6 +20x3 −8 = 0.
It is easy to show that this is a point of a local maximum of f ′′ (x). So,
3
3x(x3 + 4) 3x(x
+
4)
′′
≤
≈ 1.467889825
|f (x)| = 4(1 + x3 )3/2 4(1 + x3 )3/2 x=0.7320508076...
Therefore, taking K = 1.467889825, a = 0, b = 1, and n = 10 in the above error estimates, we
obtain
K(b − a)3
1.467889825(1 − 0)3
|ET | ≤
=
≈ 0.001223241521
12n2
12(10)2
and
1.467889825(1 − 0)3
K(b − a)3
=
≈ 0.0006116207604
|EM | ≤
24n2
24(10)2
REMARK: Note that these error estimates are bigger than the actual errors 0.000884412 and
0.000442420.
EXAMPLE: How large should we take n in order to guarantee that the Trapezoidal and MidZ2
1
dx are accurate to within 0.0001?
point Rule approximations for
x
1
6
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
EXAMPLE: How large should we take n in order to guarantee that the Trapezoidal and MidZ2
1
point Rule approximations for
dx are accurate to within 0.0001?
x
1
Solution: We saw in one of the previous examples that |f ′′(x)| ≤ 2 for 1 ≤ x ≤ 2, so we can
take K = 2, a = 1, and b = 2 in
K(b − a)3
K(b − a)3
and
|E
|
≤
M
12n2
24n2
Accuracy to within 0.0001 means that the size of the error should be less than 0.0001. Therefore,
we choose n so that
2 · 13
< 0.0001 (Trapezoidal Rule)
12n2
Solving the inequality for n, we get
1
2
=⇒ n > √
≈ 40.8
n2 >
12(0.0001)
0.0006
Thus n = 41 will ensure the desired accuracy.
|ET | ≤
For the same accuracy with the Midpoint Rule we choose n so that
2 · 13
< 0.0001
24n2
=⇒
n> √
1
≈ 29
0.0012
SIMPSON’S RULE:
Zb
f (x)dx ≈ Sn =
∆x
[f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + . . . + 2f (xn−2 ) + 4f (xn−1 ) + f (xn )]
3
a
where n is even and ∆x =
b−a
.
n
ERROR BOUND FOR SIMPSON’S RULE: Suppose that |f (4) (x)| ≤ K for a ≤ x ≤ b. If ES
is the error involved in using Simpson’s Rule, then
|ES | ≤
K(b − a)5
180n4
EXAMPLE: How large should we take n in order to guarantee that the Simpson’s Rule apZ2
proximation for x3 dx is accurate to within 0.0001?
0
7
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
EXAMPLE: How large should we take n in order to guarantee that the Simpson’s Rule apZ2
proximation for x3 dx is accurate to within 0.0001?
0
Solution: Note that
f ′ (x) = 3x2 ,
f ′′ (x) = 6x,
and f (4) (x) = 0
f ′′′ (x) = 6,
Therefore, taking K = 0 in the above error estimate, we obtain
K(b − a)5
0 · (b − a)5
=
=0
180n4
180n4
Z2
This means that Simpson’s Rule gives the exact value of x3 dx with n = 2. In fact,
|ES | ≤
0
Z2
0
which is the same as
2
x4 24
x dx =
=
=4
4 0
4
3
1 3
[0 + 4 · 13 + 23 ]
3
REMARK: One can show that if f is a polynomial of degree 3 or lower, then Simpson’s Rule
Zb
gives the exact value of f (x)dx.
a
EXAMPLE: How large should we take n in order to guarantee that the Simpson’s Rule apZ2
1
dx is accurate to within 0.0001?
proximation for
x
1
Solution: If f (x) = 1/x, then f (4) (x) = 24/x5 . Since 24/x5 is a decreasing function on [1, 2], we
have
24 24
(4)
|f (x)| = 5 ≤ 5 = 24
x
1
Therefore, we can take K = 24, a = 1, and b = 2 in
K(b − a)5
180n4
Accuracy to within 0.0001 means that the size of the error should be less than 0.0001. Therefore,
we choose n so that
24 · 15
< 0.0001
180n4
Solving the inequality for n, we get
1
24
≈ 6.04
=⇒ n > √
n4 >
4
180(0.0001)
0.00075
|ES | ≤
Thus n = 8 (n must be even) will ensure the desired accuracy.
8
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
Appendix
EXAMPLE: Let f (x) =
√
1 + x3 . Find f ′ , f ′′ , and f ′′′ .
Solution: Since f (x) = (1 + x3 )1/2 , we have
1
3x2
1
f ′ (x) = (1 + x3 )1/2−1 · (1 + x3 )′ = (1 + x3 )−1/2 · 3x2 = √
2
2
2 1 + x3
′′
f (x) =
3x2
2(1 + x3 )1/2
′
3
=
2
=
x2
(1 + x3 )1/2
′
3 (x2 )′ (1 + x3 )1/2 − x2 [(1 + x3 )1/2 ]′
·
2
[(1 + x3 )1/2 ]2
3 1/2−1
3 1/2
21
· (1 + x3 )′
3 2x(1 + x ) − x 2 (1 + x )
= ·
2
1 + x3
3 1/2
21
2x(1
+
x
)
−
x
(1 + x3 )−1/2 · 3x2
3
2
= ·
2
1 + x3
3 4
3 −1/2
3 1/2
3 2x(1 + x ) − 2 x (1 + x )
= ·
2
1 + x3
3 4
3 −1/2
3 1/2
· 2(1 + x3 )1/2
2x(1 + x ) − x (1 + x )
3
2
= ·
2
(1 + x3 ) · 2(1 + x3 )1/2
3 4
3 −1/2
3 1/2
3 1/2
· 2(1 + x3 )1/2
3 2x(1 + x ) · 2(1 + x ) − 2 x (1 + x )
= ·
2
2(1 + x3 )3/2
=
3 4x(1 + x3 ) − 3x4
·
2
2(1 + x3 )3/2
=
3 4x + 4x4 − 3x4
·
2
2(1 + x3 )3/2
=
3
4x + x4
·
2 2(1 + x3 )3/2
=
3 x(4 + x3 )
·
2 2(1 + x3 )3/2
3x(4 + x3 )
=
4(1 + x3 )3/2
9
Section 6.5 Approximate Integration
2010 Kiryl Tsishchanka
′
3x(4 + x3 )
f (x) =
4(1 + x3 )3/2
′
3 x(4 + x3 )
=
4 (1 + x3 )3/2
′′′
=
3 [x(4 + x3 )]′ (1 + x3 )3/2 − x(4 + x3 )[(1 + x3 )3/2 ]′
·
4
[(1 + x3 )3/2 ]2
3 3/2−1
′
3
3 ′
3 3/2
3 3
· (1 + x3 )′
3 [x (4 + x ) + x(4 + x ) ](1 + x ) − x(4 + x ) 2 (1 + x )
= ·
4
(1 + x3 )3
3
2
3 3/2
3 3
[1
·
(4
+
x
)
+
x
·
3x
](1
+
x
)
−
x(4
+
x
) (1 + x3 )1/2 · 3x2
3
2
= ·
3
3
4
(1 + x )
9 3
3
3 1/2
3
3
3 3/2
3 (4 + x + 3x )(1 + x ) − 2 x (4 + x )(1 + x )
= ·
4
(1 + x3 )3
9 3
3
3 1/2
3
3 3/2
3 (4 + 4x )(1 + x ) − 2 x (4 + x )(1 + x )
= ·
4
(1 + x3 )3
9 3
3
3 1/2
3
3 3/2
3 4(1 + x )(1 + x ) − 2 x (4 + x )(1 + x )
= ·
4
(1 + x3 )3
9 3
3 5/2
x (4 + x3 )(1 + x3 )1/2
4(1
+
x
)
−
3
2
= ·
4
(1 + x3 )3
9 3
3
3 1/2
3 5/2
· 2(1 + x3 )−1/2
4(1 + x ) − x (4 + x )(1 + x )
3
2
= ·
4
(1 + x3 )3 · 2(1 + x3 )−1/2
9
3 5/2
3 −1/2
− x3 (4 + x3 )(1 + x3 )1/2 · 2(1 + x3 )−1/2
3 4(1 + x ) · 2(1 + x )
2
= ·
4
2(1 + x3 )5/2
=
3 8(1 + 2x3 + x6 ) − 36x3 − 9x6
3 8(1 + x3 )2 − 9x3 (4 + x3 )
·
=
·
4
2(1 + x3 )5/2
4
2(1 + x3 )5/2
=
3 8 + 16x3 + 8x6 − 36x3 − 9x6
3 8 − 20x3 − x6
·
=
·
4
2(1 + x3 )5/2
4 2(1 + x3 )5/2
=
3(x6 + 20x3 − 8)
3(8 − 20x3 − x6 )
=
−
8(1 + x3 )5/2
8(1 + x3 )5/2
10