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Approximate Integration Carmen Bruni February 13, 2012
Approximate Integration
Carmen Bruni
February 13, 2012
Definite Integrals
Sometimes its impossible to compute the indefinite integral of afunction.
For example, if
f (x) =√
1 + x3 sin(6x) + 14
then ∫f (x)dx
has no nice formula.However, we should be able to compute or atleast approximate definite integrals of f (x), for example∫ 5
1f (x)dx .
After all, definite integrals just represent area.
Definite Integrals
Sometimes its impossible to compute the indefinite integral of afunction. For example, if
f (x) =√
1 + x3 sin(6x) + 14
then
∫f (x)dx
has no nice formula.However, we should be able to compute or atleast approximate definite integrals of f (x), for example∫ 5
1f (x)dx .
After all, definite integrals just represent area.
Definite Integrals
Sometimes its impossible to compute the indefinite integral of afunction. For example, if
f (x) =√
1 + x3 sin(6x) + 14
then ∫f (x)dx
has no nice formula.
However, we should be able to compute or atleast approximate definite integrals of f (x), for example∫ 5
1f (x)dx .
After all, definite integrals just represent area.
Definite Integrals
Sometimes its impossible to compute the indefinite integral of afunction. For example, if
f (x) =√
1 + x3 sin(6x) + 14
then ∫f (x)dx
has no nice formula.However, we should be able to compute or atleast approximate definite integrals of f (x), for example∫ 5
1f (x)dx .
After all, definite integrals just represent area.
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Definite Integral of f (x) =√
1 + x3 sin(6x) + 14
How do we approximate definite integrals?
We have already seen 3 ways. Let x0, x1, .., xn be a partition of[a, b] and let ∆x = b−an .
1. Left Endpoints Ln =n∑
i=0
f (xi−1)∆x .
2. Right Endpoints Rn =n∑
i=0
f (xi )∆x .
3. Midpoint Rule Mn =n∑
i=0
f (x̄i )∆x , where x̄i =xi−1+xi
2 .
We will learn two additional ways today
1. Trapezoid Rule
2. Simpson’s Rule
Left Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14Let’s try each of these methods with our function f (x) on [1, 5]and using a partition of four subintervals. In these cases,∆x = 5−14 = 1.
Left Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14Let’s try each of these methods with our function f (x) on [1, 5]and using a partition of four subintervals. In these cases,∆x = 5−14 = 1.
Left Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ L4 = f (1)∆x + f (2)∆x + f (3)∆x + f (4)∆x
= (√
1 + (1)3 sin(6(1)) + 14)(1)
+(√
1 + (2)3 sin(6(2)) + 14)(1)
+(√
1 + (3)3 sin(6(3)) + 14)(1)
+(√
1 + (4)3 sin(6(4)) + 14)(1)
= 42.72027090
Left Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ L4 = f (1)∆x + f (2)∆x + f (3)∆x + f (4)∆x
= (√
1 + (1)3 sin(6(1)) + 14)(1)
+(√
1 + (2)3 sin(6(2)) + 14)(1)
+(√
1 + (3)3 sin(6(3)) + 14)(1)
+(√
1 + (4)3 sin(6(4)) + 14)(1)
= 42.72027090
Left Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ L4 = f (1)∆x + f (2)∆x + f (3)∆x + f (4)∆x
= (√
1 + (1)3 sin(6(1)) + 14)(1)
+(√
1 + (2)3 sin(6(2)) + 14)(1)
+(√
1 + (3)3 sin(6(3)) + 14)(1)
+(√
1 + (4)3 sin(6(4)) + 14)(1)
= 42.72027090
Right Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Right endpoint rule now.
Right Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Right endpoint rule now.
Right Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ R4 = f (2)∆x + f (3)∆x + f (4)∆x + f (5)∆x
= (√
1 + (2)3 sin(6(2)) + 14)(1)
+(√
1 + (3)3 sin(6(3)) + 14)(1)
+(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1)
= 32.02479662
Right Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ R4 = f (2)∆x + f (3)∆x + f (4)∆x + f (5)∆x
= (√
1 + (2)3 sin(6(2)) + 14)(1)
+(√
1 + (3)3 sin(6(3)) + 14)(1)
+(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1)
= 32.02479662
Right Endpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ R4 = f (2)∆x + f (3)∆x + f (4)∆x + f (5)∆x
= (√
1 + (2)3 sin(6(2)) + 14)(1)
+(√
1 + (3)3 sin(6(3)) + 14)(1)
+(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1)
= 32.02479662
Midpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Midpoint rule now.
Midpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Midpoint rule now.
Midpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ M4 = f (1.5)∆x + f (2.5)∆x + f (3.5)∆x + f (4.5)∆x
= (√
1 + (1.5)3 sin(6(1.5)) + 14)(1)
+(√
1 + (2.5)3 sin(6(2.5)) + 14)(1)
+(√
1 + (3.5)3 sin(6(3.5)) + 14)(1)
+(√
1 + (4.5)3 sin(6(4.5)) + 14)(1)
= 74.23479834
Midpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ M4 = f (1.5)∆x + f (2.5)∆x + f (3.5)∆x + f (4.5)∆x
= (√
1 + (1.5)3 sin(6(1.5)) + 14)(1)
+(√
1 + (2.5)3 sin(6(2.5)) + 14)(1)
+(√
1 + (3.5)3 sin(6(3.5)) + 14)(1)
+(√
1 + (4.5)3 sin(6(4.5)) + 14)(1)
= 74.23479834
Midpoint Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ M4 = f (1.5)∆x + f (2.5)∆x + f (3.5)∆x + f (4.5)∆x
= (√
1 + (1.5)3 sin(6(1.5)) + 14)(1)
+(√
1 + (2.5)3 sin(6(2.5)) + 14)(1)
+(√
1 + (3.5)3 sin(6(3.5)) + 14)(1)
+(√
1 + (4.5)3 sin(6(4.5)) + 14)(1)
= 74.23479834
Other Rules? The Trapezoid Rule.
How do we get new rules?
One way is to take combinations of theabove rules. For example, lets try taking half the left endpoint ruleand half the right endpoint rule. This gives us the trapezoid rule.
Tn =1
2Ln +
1
2Rn
=1
2
n∑i=0
f (xi−1)∆x +1
2
n∑i=0
f (xi−1)∆x
=1
2
((f (x0)∆x + f (x1)∆x + ... + f (xn−1)∆x)
+(f (x1)∆x + ... + f (xn−1)∆x + f (xn)∆x))
=∆x
2(f (x0) + 2f (x1) + ... + 2f (xn−1) + f (xn))
Other Rules? The Trapezoid Rule.
How do we get new rules? One way is to take combinations of theabove rules. For example, lets try taking half the left endpoint ruleand half the right endpoint rule. This gives us the trapezoid rule.
Tn =1
2Ln +
1
2Rn
=1
2
n∑i=0
f (xi−1)∆x +1
2
n∑i=0
f (xi−1)∆x
=1
2
((f (x0)∆x + f (x1)∆x + ... + f (xn−1)∆x)
+(f (x1)∆x + ... + f (xn−1)∆x + f (xn)∆x))
=∆x
2(f (x0) + 2f (x1) + ... + 2f (xn−1) + f (xn))
Other Rules? The Trapezoid Rule.
How do we get new rules? One way is to take combinations of theabove rules. For example, lets try taking half the left endpoint ruleand half the right endpoint rule. This gives us the trapezoid rule.
Tn =1
2Ln +
1
2Rn
=1
2
n∑i=0
f (xi−1)∆x +1
2
n∑i=0
f (xi−1)∆x
=1
2
((f (x0)∆x + f (x1)∆x + ... + f (xn−1)∆x)
+(f (x1)∆x + ... + f (xn−1)∆x + f (xn)∆x))
=∆x
2(f (x0) + 2f (x1) + ... + 2f (xn−1) + f (xn))
Other Rules? The Trapezoid Rule.
How do we get new rules? One way is to take combinations of theabove rules. For example, lets try taking half the left endpoint ruleand half the right endpoint rule. This gives us the trapezoid rule.
Tn =1
2Ln +
1
2Rn
=1
2
n∑i=0
f (xi−1)∆x +1
2
n∑i=0
f (xi−1)∆x
=1
2
((f (x0)∆x + f (x1)∆x + ... + f (xn−1)∆x)
+(f (x1)∆x + ... + f (xn−1)∆x + f (xn)∆x))
=∆x
2(f (x0) + 2f (x1) + ... + 2f (xn−1) + f (xn))
Other Rules? The Trapezoid Rule.
How do we get new rules? One way is to take combinations of theabove rules. For example, lets try taking half the left endpoint ruleand half the right endpoint rule. This gives us the trapezoid rule.
Tn =1
2Ln +
1
2Rn
=1
2
n∑i=0
f (xi−1)∆x +1
2
n∑i=0
f (xi−1)∆x
=1
2
((f (x0)∆x + f (x1)∆x + ... + f (xn−1)∆x)
+(f (x1)∆x + ... + f (xn−1)∆x + f (xn)∆x))
=∆x
2(f (x0) + 2f (x1) + ... + 2f (xn−1) + f (xn))
Trapezoid Rule for f (x) =√
1 + x3 sin(6x) + 14
Why is it called the trapezoid rule? Well,
Trapezoid Rule for f (x) =√
1 + x3 sin(6x) + 14
Why is it called the trapezoid rule? Well,
Trapezoid Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ T4 =∆x
2(f (x0) + 2f (x1) + 2f (x2) + 2f (x3) + f (x4))
= 0.5(
(√
1 + (1)3 sin(6(1)) + 14)(1)
+2(√
1 + (2)3 sin(6(2)) + 14)(1)
+2(√
1 + (3)3 sin(6(3)) + 14)(1)
+2(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1))
= 37.37253376
Trapezoid Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ T4 =∆x
2(f (x0) + 2f (x1) + 2f (x2) + 2f (x3) + f (x4))
= 0.5(
(√
1 + (1)3 sin(6(1)) + 14)(1)
+2(√
1 + (2)3 sin(6(2)) + 14)(1)
+2(√
1 + (3)3 sin(6(3)) + 14)(1)
+2(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1))
= 37.37253376
Trapezoid Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ T4 =∆x
2(f (x0) + 2f (x1) + 2f (x2) + 2f (x3) + f (x4))
= 0.5(
(√
1 + (1)3 sin(6(1)) + 14)(1)
+2(√
1 + (2)3 sin(6(2)) + 14)(1)
+2(√
1 + (3)3 sin(6(3)) + 14)(1)
+2(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1))
= 37.37253376
How good are these rules?
How close do these rules get to the actual answer?
It turns outthat the left and right endpoint rules do not give us a goodestimate on how far we are from the actual answer. The midpointand trapezoid rules however do lend themselves to good errorbounds. To describe this, we define the error of the trapezoid ruleto be
ET =
∫ ba
f (x)dx − Tn
and the error of the midpoint rule to be
EM =
∫ ba
f (x)dx −Mn.
Note that we could only compute the error term exactly if we knewthe value of the integral already (and if we knew that then what’sthe point of doing this!) However, it turns out that we can boundthe error.
How good are these rules?
How close do these rules get to the actual answer? It turns outthat the left and right endpoint rules do not give us a goodestimate on how far we are from the actual answer. The midpointand trapezoid rules however do lend themselves to good errorbounds.
To describe this, we define the error of the trapezoid ruleto be
ET =
∫ ba
f (x)dx − Tn
and the error of the midpoint rule to be
EM =
∫ ba
f (x)dx −Mn.
Note that we could only compute the error term exactly if we knewthe value of the integral already (and if we knew that then what’sthe point of doing this!) However, it turns out that we can boundthe error.
How good are these rules?
How close do these rules get to the actual answer? It turns outthat the left and right endpoint rules do not give us a goodestimate on how far we are from the actual answer. The midpointand trapezoid rules however do lend themselves to good errorbounds. To describe this, we define the error of the trapezoid ruleto be
ET =
∫ ba
f (x)dx − Tn
and the error of the midpoint rule to be
EM =
∫ ba
f (x)dx −Mn.
Note that we could only compute the error term exactly if we knewthe value of the integral already (and if we knew that then what’sthe point of doing this!) However, it turns out that we can boundthe error.
How good are these rules?
How close do these rules get to the actual answer? It turns outthat the left and right endpoint rules do not give us a goodestimate on how far we are from the actual answer. The midpointand trapezoid rules however do lend themselves to good errorbounds. To describe this, we define the error of the trapezoid ruleto be
ET =
∫ ba
f (x)dx − Tn
and the error of the midpoint rule to be
EM =
∫ ba
f (x)dx −Mn.
Note that we could only compute the error term exactly if we knewthe value of the integral already (and if we knew that then what’sthe point of doing this!) However, it turns out that we can boundthe error.
How good are these rules?
How close do these rules get to the actual answer? It turns outthat the left and right endpoint rules do not give us a goodestimate on how far we are from the actual answer. The midpointand trapezoid rules however do lend themselves to good errorbounds. To describe this, we define the error of the trapezoid ruleto be
ET =
∫ ba
f (x)dx − Tn
and the error of the midpoint rule to be
EM =
∫ ba
f (x)dx −Mn.
Note that we could only compute the error term exactly if we knewthe value of the integral already (and if we knew that then what’sthe point of doing this!) However, it turns out that we can boundthe error.
A Bound on the Error
Theorem: Suppose that |f ′′(x)| ≤ K for all x ∈ [a, b] and forsome constant K ∈ R. If ET and EM are the errors in thetrapezoid and midpoint rules respectively, then
|ET | ≤K (b − a)3
12n2
and
|EM | ≤K (b − a)3
24n2.
Notice that the bound for EM is twice as good as the bound forET .
A Bound on the Error
Theorem: Suppose that |f ′′(x)| ≤ K for all x ∈ [a, b] and forsome constant K ∈ R. If ET and EM are the errors in thetrapezoid and midpoint rules respectively, then
|ET | ≤K (b − a)3
12n2
and
|EM | ≤K (b − a)3
24n2.
Notice that the bound for EM is twice as good as the bound forET .
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?
Solution: Set f (x) = e−x2. Then f ′(x) = −2xe−x2 and
f ′′(x) = −2e−x2 + 4x2e−x2 = e−x2(4x2 − 2).
I claim that this function is increasing form 0 to 1. To see this,notice that the third derivative is
f ′′′(x) = e−x2(12x − 8x3) = 4xe−x2(3− 2x2)
and this function is greater than or equal to 0 between 0 and 1.Hence the derivative is positive on [0, 1] and so the function isincreasing there. Hence the maximum value of f ′′(x) on [0, 1]occurs at x = 1 which gives the value
K = f ′′(1) = e−(1)2(4(1)2 − 2) = 2
e= 0.7357588824
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?
Solution: Set f (x) = e−x2.
Then f ′(x) = −2xe−x2 and
f ′′(x) = −2e−x2 + 4x2e−x2 = e−x2(4x2 − 2).
I claim that this function is increasing form 0 to 1. To see this,notice that the third derivative is
f ′′′(x) = e−x2(12x − 8x3) = 4xe−x2(3− 2x2)
and this function is greater than or equal to 0 between 0 and 1.Hence the derivative is positive on [0, 1] and so the function isincreasing there. Hence the maximum value of f ′′(x) on [0, 1]occurs at x = 1 which gives the value
K = f ′′(1) = e−(1)2(4(1)2 − 2) = 2
e= 0.7357588824
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?
Solution: Set f (x) = e−x2. Then f ′(x) = −2xe−x2 and
f ′′(x) = −2e−x2 + 4x2e−x2 = e−x2(4x2 − 2).
I claim that this function is increasing form 0 to 1. To see this,notice that the third derivative is
f ′′′(x) = e−x2(12x − 8x3) = 4xe−x2(3− 2x2)
and this function is greater than or equal to 0 between 0 and 1.Hence the derivative is positive on [0, 1] and so the function isincreasing there. Hence the maximum value of f ′′(x) on [0, 1]occurs at x = 1 which gives the value
K = f ′′(1) = e−(1)2(4(1)2 − 2) = 2
e= 0.7357588824
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?
Solution: Set f (x) = e−x2. Then f ′(x) = −2xe−x2 and
f ′′(x) = −2e−x2 + 4x2e−x2 = e−x2(4x2 − 2).
I claim that this function is increasing form 0 to 1. To see this,notice that the third derivative is
f ′′′(x) = e−x2(12x − 8x3) = 4xe−x2(3− 2x2)
and this function is greater than or equal to 0 between 0 and 1.Hence the derivative is positive on [0, 1] and so the function isincreasing there. Hence the maximum value of f ′′(x) on [0, 1]occurs at x = 1 which gives the value
K = f ′′(1) = e−(1)2(4(1)2 − 2) = 2
e= 0.7357588824
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?
Solution: Set f (x) = e−x2. Then f ′(x) = −2xe−x2 and
f ′′(x) = −2e−x2 + 4x2e−x2 = e−x2(4x2 − 2).
I claim that this function is increasing form 0 to 1. To see this,notice that the third derivative is
f ′′′(x) = e−x2(12x − 8x3) = 4xe−x2(3− 2x2)
and this function is greater than or equal to 0 between 0 and 1.
Hence the derivative is positive on [0, 1] and so the function isincreasing there. Hence the maximum value of f ′′(x) on [0, 1]occurs at x = 1 which gives the value
K = f ′′(1) = e−(1)2(4(1)2 − 2) = 2
e= 0.7357588824
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?
Solution: Set f (x) = e−x2. Then f ′(x) = −2xe−x2 and
f ′′(x) = −2e−x2 + 4x2e−x2 = e−x2(4x2 − 2).
I claim that this function is increasing form 0 to 1. To see this,notice that the third derivative is
f ′′′(x) = e−x2(12x − 8x3) = 4xe−x2(3− 2x2)
and this function is greater than or equal to 0 between 0 and 1.Hence the derivative is positive on [0, 1] and so the function isincreasing there. Hence the maximum value of f ′′(x) on [0, 1]occurs at x = 1 which gives the value
K = f ′′(1) = e−(1)2(4(1)2 − 2) = 2
e= 0.7357588824
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about
with the midpoint rule?Solution: With K = 0.7357588824, we use the error boundingtheorem and note that
ET ≤ 0.001 wheneverK (b − a)3
12n2≤ 0.001.
Plugging in our K , a = 0 and b = 1 values and solving for n, wesee that we need n at least as large as
K (b − a)3
12n2≤ 0.001 ⇒ (0.7357588824)(1− 0)3 ≤ 12(0.001)n2
⇒
√0.7357588824
12(0.001)≤ n
⇒ 7.830277147 ≤ n
So we need n to be at least 8.
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about
with the midpoint rule?Solution: With K = 0.7357588824, we use the error boundingtheorem and note that
ET ≤ 0.001 wheneverK (b − a)3
12n2≤ 0.001.
Plugging in our K , a = 0 and b = 1 values and solving for n, wesee that we need n at least as large as
K (b − a)3
12n2≤ 0.001 ⇒ (0.7357588824)(1− 0)3 ≤ 12(0.001)n2
⇒
√0.7357588824
12(0.001)≤ n
⇒ 7.830277147 ≤ n
So we need n to be at least 8.
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?Solution: For the midpoint rule, we use the error boundingtheorem and note that
EM ≤ 0.001 wheneverK (b − a)3
24n2≤ 0.001.
Plugging in our K = 0.7357588824, a = 0 and b = 1 values andsolving for n, we see that we need n at least as large as
K (b − a)3
24n2≤ 0.001 ⇒ (0.7357588824)(1− 0)3 ≤ 24(0.001)n2
⇒
√0.7357588824
24(0.001)≤ n
⇒ 5.536842069 ≤ n
So we need n to be at least 6. Notice that this n is smaller than theone needed for the trapezoid rule because its a better estimation.
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?Solution: For the midpoint rule, we use the error boundingtheorem and note that
EM ≤ 0.001 wheneverK (b − a)3
24n2≤ 0.001.
Plugging in our K = 0.7357588824, a = 0 and b = 1 values andsolving for n, we see that we need n at least as large as
K (b − a)3
24n2≤ 0.001 ⇒ (0.7357588824)(1− 0)3 ≤ 24(0.001)n2
⇒
√0.7357588824
24(0.001)≤ n
⇒ 5.536842069 ≤ n
So we need n to be at least 6.
Notice that this n is smaller than theone needed for the trapezoid rule because its a better estimation.
Example Question using the error boundQuestion: How large does n have to be in order to estimate∫ 10 e−x2 dx within 0.001 using the trapezoid rule? How about with
the midpoint rule?Solution: For the midpoint rule, we use the error boundingtheorem and note that
EM ≤ 0.001 wheneverK (b − a)3
24n2≤ 0.001.
Plugging in our K = 0.7357588824, a = 0 and b = 1 values andsolving for n, we see that we need n at least as large as
K (b − a)3
24n2≤ 0.001 ⇒ (0.7357588824)(1− 0)3 ≤ 24(0.001)n2
⇒
√0.7357588824
24(0.001)≤ n
⇒ 5.536842069 ≤ n
So we need n to be at least 6. Notice that this n is smaller than theone needed for the trapezoid rule because its a better estimation.
Other Rules? Simpson’s Rule.
Named after English mathematician Thomas Simpson(1710-1761). For this rule, we can only apply it with an evennumber of sub intervals. We try to combine the left, right, andmidpoint rules. We start by trying a third of each.
1
3Ln +
1
3Rn +
1
3Mn =
1
3(Ln + Rn) +
1
3Mn =
2
3Tn +
1
3Mn
Remember the error bounds? We said that Mn was twice as goodas Tn and so these fractions are the wrong way. To change this,we weigh the Mn piece twice as much. This gives
S2n =1
6Ln +
1
6Rn +
2
3Mn =
1
3Tn +
2
3Mn.
Other Rules? Simpson’s Rule.
Named after English mathematician Thomas Simpson(1710-1761). For this rule, we can only apply it with an evennumber of sub intervals. We try to combine the left, right, andmidpoint rules. We start by trying a third of each.
1
3Ln +
1
3Rn +
1
3Mn =
1
3(Ln + Rn) +
1
3Mn =
2
3Tn +
1
3Mn
Remember the error bounds? We said that Mn was twice as goodas Tn and so these fractions are the wrong way. To change this,we weigh the Mn piece twice as much. This gives
S2n =1
6Ln +
1
6Rn +
2
3Mn =
1
3Tn +
2
3Mn.
Simpson’s Rule expanded
Let’s expand this rule. We need a diagram.
So notice that yi = x2i , ȳi = x2i−1 and
∆y =b − an
= 2b − a
2n= 2∆x
Simpson’s Rule expanded
Let’s expand this rule. We need a diagram.
So notice that yi = x2i , ȳi = x2i−1 and
∆y =b − an
= 2b − a
2n= 2∆x
A Formula for Simpson’s RuleHence,
S2n =1
3Tn +
2
3Mn
=1
3· ∆y
2(f (y0) + 2f (y1) + ... + 2f (yn−1) + f (yn))
+2
3·∆y (f (ȳ1) + f (ȳ2) + ... + f (ȳn))
=∆x
3
(f (x2(0)) + 2f (x2(1)) + ... + 2f (x2(n−1)) + f (x2(n))
)+
4∆x
3
(f (x2(1)−1) + f (x2(2)−1) + ... + f (x2(n)−1)
)=
∆x
3
((f (x0) + 2f (x2) + ... + 2f (x2n−2) + f (x2n))
+(4f (x1) + 4f (x3) + ... + 4f (x2n−1)))
=∆x
3
(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + ...
+2f (x2n−2) + 4f (x2n−1) + f (x2n))
A Formula for Simpson’s RuleHence,
S2n =1
3Tn +
2
3Mn
=1
3· ∆y
2(f (y0) + 2f (y1) + ... + 2f (yn−1) + f (yn))
+2
3·∆y (f (ȳ1) + f (ȳ2) + ... + f (ȳn))
=∆x
3
(f (x2(0)) + 2f (x2(1)) + ... + 2f (x2(n−1)) + f (x2(n))
)+
4∆x
3
(f (x2(1)−1) + f (x2(2)−1) + ... + f (x2(n)−1)
)=
∆x
3
((f (x0) + 2f (x2) + ... + 2f (x2n−2) + f (x2n))
+(4f (x1) + 4f (x3) + ... + 4f (x2n−1)))
=∆x
3
(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + ...
+2f (x2n−2) + 4f (x2n−1) + f (x2n))
A Formula for Simpson’s RuleHence,
S2n =1
3Tn +
2
3Mn
=1
3· ∆y
2(f (y0) + 2f (y1) + ... + 2f (yn−1) + f (yn))
+2
3·∆y (f (ȳ1) + f (ȳ2) + ... + f (ȳn))
=∆x
3
(f (x2(0)) + 2f (x2(1)) + ... + 2f (x2(n−1)) + f (x2(n))
)+
4∆x
3
(f (x2(1)−1) + f (x2(2)−1) + ... + f (x2(n)−1)
)
=∆x
3
((f (x0) + 2f (x2) + ... + 2f (x2n−2) + f (x2n))
+(4f (x1) + 4f (x3) + ... + 4f (x2n−1)))
=∆x
3
(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + ...
+2f (x2n−2) + 4f (x2n−1) + f (x2n))
A Formula for Simpson’s RuleHence,
S2n =1
3Tn +
2
3Mn
=1
3· ∆y
2(f (y0) + 2f (y1) + ... + 2f (yn−1) + f (yn))
+2
3·∆y (f (ȳ1) + f (ȳ2) + ... + f (ȳn))
=∆x
3
(f (x2(0)) + 2f (x2(1)) + ... + 2f (x2(n−1)) + f (x2(n))
)+
4∆x
3
(f (x2(1)−1) + f (x2(2)−1) + ... + f (x2(n)−1)
)=
∆x
3
((f (x0) + 2f (x2) + ... + 2f (x2n−2) + f (x2n))
+(4f (x1) + 4f (x3) + ... + 4f (x2n−1)))
=∆x
3
(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + ...
+2f (x2n−2) + 4f (x2n−1) + f (x2n))
A Formula for Simpson’s RuleHence,
S2n =1
3Tn +
2
3Mn
=1
3· ∆y
2(f (y0) + 2f (y1) + ... + 2f (yn−1) + f (yn))
+2
3·∆y (f (ȳ1) + f (ȳ2) + ... + f (ȳn))
=∆x
3
(f (x2(0)) + 2f (x2(1)) + ... + 2f (x2(n−1)) + f (x2(n))
)+
4∆x
3
(f (x2(1)−1) + f (x2(2)−1) + ... + f (x2(n)−1)
)=
∆x
3
((f (x0) + 2f (x2) + ... + 2f (x2n−2) + f (x2n))
+(4f (x1) + 4f (x3) + ... + 4f (x2n−1)))
=∆x
3
(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + ...
+2f (x2n−2) + 4f (x2n−1) + f (x2n))
Simpson’s Rule for f (x) =√
1 + x3 sin(6x) + 14
Our last example,
Simpson’s Rule for f (x) =√
1 + x3 sin(6x) + 14
Our last example,
Simpson’s Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ S4 =∆x
3(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + f (x4))
= 13
((√
1 + (1)3 sin(6(1)) + 14)(1)
+4(√
1 + (2)3 sin(6(2)) + 14)(1)
+2(√
1 + (3)3 sin(6(3)) + 14)(1)
+4(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1))
= 37.04120590
Simpson’s Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ S4 =∆x
3(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + f (x4))
= 13
((√
1 + (1)3 sin(6(1)) + 14)(1)
+4(√
1 + (2)3 sin(6(2)) + 14)(1)
+2(√
1 + (3)3 sin(6(3)) + 14)(1)
+4(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1))
= 37.04120590
Simpson’s Rule for f (x) =√
1 + x3 sin(6x) + 14
Using the formula, we see that∫ 51
f (x)dx ≈ S4 =∆x
3(f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + f (x4))
= 13
((√
1 + (1)3 sin(6(1)) + 14)(1)
+4(√
1 + (2)3 sin(6(2)) + 14)(1)
+2(√
1 + (3)3 sin(6(3)) + 14)(1)
+4(√
1 + (4)3 sin(6(4)) + 14)(1)
+(√
1 + (5)3 sin(6(5)) + 14)(1))
= 37.04120590
A Bound on the Error
Let m be an even integer. As before, let
ES =
∫ ba
f (x)dx − Sm.
Then
Theorem: Suppose that |f ′′′′(x)| ≤ K for all x ∈ [a, b] and forsome constant K ∈ R. Then
|ES | ≤K (b − a)5
180m4.
A Bound on the Error
Let m be an even integer. As before, let
ES =
∫ ba
f (x)dx − Sm.
Then
Theorem: Suppose that |f ′′′′(x)| ≤ K for all x ∈ [a, b] and forsome constant K ∈ R. Then
|ES | ≤K (b − a)5
180m4.
Definite Integrals
