contents · chapter 7 notes, stewart 7e chalmeta 7.1 integration by parts introduction if the...
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Chapter 7 Notes, Stewart 7e Chalmeta
Contents
7.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27.2 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
7.2.1 Evaluating
∫
sinm x cosn(x)dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
7.2.2 Evaluating
∫
tanm x secn xdx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
7.2.3 Evaluating
∫
tanxdx,
∫
cotxdx,
∫
secxdx, and
∫
cscxdx . . . . . . . . . . . . . 12
7.3 Trigonometric Substitution for Integrals Involving√a2 − x2,
√a2 + x2,
√x2 − a2 . . . . . . 14
7.4 Integration of Rational Functions by Partial Fractions . . . . . . . . . . . . . . . . . . . . . 177.4.1 Reducing an Improper Fraction (long division) . . . . . . . . . . . . . . . . . . . . . 177.4.2 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187.4.3 Rationalizing substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207.4.4 Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
7.5 Strategy for Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.5.1 Basic Integration Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.5.2 Procedures for matching integrals to basic formulas . . . . . . . . . . . . . . . . . . . 25
7.6 Integral Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297.7 Approximate Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
7.7.1 Midpoint Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337.7.2 Trapezoidal Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347.7.3 Simpson’s / Parabolic Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1
Chapter 7 Notes, Stewart 7e Chalmeta
7.1 Integration by Parts
Introduction
If the techniques/formulas we introduced earlier do not work then there is another technique that you
can use: Integration by Parts. It is a way of simplifying integrals of the form
∫
f(x)g(x)dx in which f(x)
can be differentiated repeatedly and g(x) can be integrated repeatedly without difficulty.
The Formula
A. Derivation of the Formula The formula for integration by parts comes from the Product Rule.
d
dx(uv) = u
dv
dx+ v
du
dxd(uv) = udv + v du (differential form)
udv = d(uv)− v du (Solve for udv )∫
udv = uv −∫
v du (integrating both sides)
B. Note: The integration by parts expresses one integral,∫
v du in terms of another,∫
udv. The idea isthat with a proper choice of u and v, the second integral is easier to integrate than the original. Youmay have to use this technique several times before you reach an integral that you can integrate easily.
C. The Integration-by-Parts Formula
1. Indefinite Integrals
If u(x) and v(x) are functions of x and have continuous derivatives, then
∫
udv = uv −∫
v du
2. Definite Integrals∫ b
audv = uv
∣
∣
∣
∣
b
a
−∫ b
av du
D. How to pick u and dv
When deciding on your choice for u and dv, use the acronym: ILATE with the higher one having thepriority for u.
I -inverse trig functions
L -logarithmic functions
A -algebraic functions
T -trigonometric functions
E -exponential functions
2
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.1.1.
∫
3xe2x dx
Example 7.1.2.
∫
lnxdx
Example 7.1.3.
∫ e
1x2 lnxdx
3
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.1.6.
∫
t4 sin (2t)dt
Example 7.1.7.
∫
ex cos (4x)dx
Example 7.1.8. Use the reduction formula
∫
secn udu =1
n− 1tanu secn−2 u+
n− 2
n− 1
∫
secn−2 udu
∫
sec4 y dy
5
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.1.9.
∫
t3 cos (t2)dt
Example 7.1.10. Find the area of the region enclosed by the curve y = x cosx and the x-axis fromx = π
2 to x = 3π2
x
y
ππ2
3π2
y = x cos(x)
0
−1
−2
−3
6
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.1.11. Find the volume of the solid generated by revolving the region in the 1st quadrantbounded by the axes, the curve y = ex and the line x = ln 2 about the line x = ln 2.
x
y
ln 2
y = ex, x = ln 2, x = 0, y = 0
1
2
1
A Summary of Common Integrals Using Integration by Parts:∫
xneax dx
∫
xn sin(ax)dx
∫
xn cos(ax)dx
∫
xn lnxdx
∫
xn sin−1(ax)dx
∫
xn tan−1(ax)dx
∫
eax sin(bx)dx
∫
eax cos(bx)dx
7
Chapter 7 Notes, Stewart 7e Chalmeta
7.2 Trigonometric Integrals
7.2.1 Evaluating
∫
sinm x cosn(x)dx
A. If the power of cosine is odd (n = 2k+ 1), save one cosine factor and use cos2 x = 1− sin2 x to expressthe remaining factors in terms of sine:
∫
sinm x cos2k+1 xdx =
∫
sinm x cos2k x cosxdx
=
∫
sinm x(
cos2 x)k
cosxdx
=
∫
sinm x(
1− sin2 x)k
cosxdx
Then substitute u = sinx.
B. If the power of sine is odd (m = 2k+1), save one sine factor and use sin2 x = 1− cos2 x to express theremaining factors in terms of cosine:
∫
sin2k+1 x cosn xdx =
∫
sin2k x cosn x sinxdx
=
∫
(
sin2 x)k
cosn x sinxdx
=
∫
(
1− cos2 x)k
cosn x sinxdx
Then substitute u = cosx.
C. If the powers of both sine and cosine are odd, either 1 or 2 can be used.
D. If the powers of both sine and cosine are even, use the half-angle identities
sin2 x =1
2[1− cos (2x)] and cos2 x =
1
2[1 + cos (2x)]
E. It is sometimes helpful to use the identity sinx cosx = 12 sin(2x).
Example 7.2.1.
∫
sin θ cos4(θ)dθ
8
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.2.2.
∫
sin3 φdφ
Example 7.2.3.
∫
sin3√x cos3
√x√
xdx
9
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.2.4. Find the area bounded by the curve of y = sin4(3x) and the x-axis from x = 0 to x = π3
x
y
π3
y = sin4(3x)
1
7.2.2 Evaluating
∫
tanm x secn xdx
A. If the power of secant is even (n = 2k), save one factor of sec2 x and use sec2 x = 1+ tan2 x to expressthe remaining factors in terms of tangent:
∫
tanm x sec2k xdx =
∫
tanm x sec2k−2 x sec2 xdx
=
∫
tanm x(
sec2 x)k−1
sec2 xdx
=
∫
tanm x(
1 + tan2 x)k−1
sec2 xdx
Then substitute u = tanx.
10
Chapter 7 Notes, Stewart 7e Chalmeta
B. If the power of tangent is odd (m = 2k + 1), save one factor of secx tanx and use tan2 x = sec2 x− 1to express the remaining factors in terms of secant:
∫
tan2k+1 x secn xdx =
∫
tan2k x secn−1 x secx tanxdx
=
∫
(
tan2 x)k
secn−1 x secx tanxdx
=
∫
(
sec2 x− 1)k
secn−1 x secx tanxdx
Then substitute u = secx.
Example 7.2.5.
∫
tan6 y dy
Example 7.2.6.
∫ π/6
0tan θ sec3 θ dθ
11
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.2.7. Find the volume of the solid generated by revolving the region bounded by the curve of
y = sec2(x2) tan2(x2), the x-axis and the line x =√π2 about the y-axis.
x
y
√π2
y = sec2(x2) tan2(x2)
1
2
7.2.3 Evaluating
∫
tanxdx,
∫
cotxdx,
∫
secxdx, and
∫
cscxdx
1. When evaluating
∫
tanxdx or
∫
cotxdx rewrite the integrand in terms of sine and cosine and then
use u-substitution.
2. When evaluating
∫
secxdx multiply the integrand by the form of one,secx+ tanx
secx+ tanx, and then use
u-substitution.
3. When evaluating
∫
cscxdx multiply the integrand by the form of one,cscx+ cotx
cscx+ cotx, and then use
u-substitution.
12
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.2.8.
∫ π/2
π/4cot θ dθ
Example 7.2.9.
∫
secxdx
Integral Formulas for tan(x), cot(x), sec(x), csc(x)
1.
∫
tanudu = − ln | cosu|+ C = ln | secu|+ C
2.
∫
cotudu = ln | sinu|+ C = − ln | cscu|+ C
3.
∫
secudu = ln | secu+ tanu|+ C
4.
∫
cscudu = − ln | cscu+ cotu|+ C
13
Chapter 7 Notes, Stewart 7e Chalmeta
7.3 Trigonometric Substitution for Integrals Involving√a2 − x
2,√a2 + x
2,√x2 − a
2
Introduction
Consider the integral∫
x√a2 − x2 dx. The substitution of u = a2 − x2 will allow us to integrate this
integral.Now consider
∫ √a2 − x2 dx. We have integrated this type of integral using the area of either a half or
quarter of a circle depending on the limits of integration. What if we wanted to evaluate this indefiniteintegral or the definite integral in which the limits do not define either a half or quarter of a circle? Wewill evaluate this type of integral using a type of substitution called inverse substitution.
Trigonometric Substitution for Integrals Involving√a2 − x2,
√a2 + x2,
√x2 − a2
Radical Substitution Restrictions on θ Identiy
√a2 − x2 x = a sin θ −π
2 ≤ θ ≤ π2 1− sin2 θ = cos2 θ
√a2 + x2 x = a tan θ −π
2 ≤ θ ≤ π2 1 + tan2 θ = sec2 θ
√x2 − a2 x = a sec θ 0 ≤ θ ≤ π, θ 6= π
2 sec2 θ − 1 = tan2 θ
Example 7.3.1. Evaluate∫ √
a2 − x2 dx
14
Chapter 7 Notes, Stewart 7e Chalmeta
Simplifications of the trigonometric substitutions
1.√a2 − x2 =
√
a2 − a2 sin2 θ =√a2 cos2 θ = |a cos θ| = a cos θ
2.√a2 + x2 =
√a2 + a2 tan2 θ =
√a2 sec2 θ = |a sec θ| = a sec θ
3.√x2 − a2 =
√a2 sec2 θ − a2 =
√a2 tan2 θ = |a tan θ| = a tan θ
Example 7.3.2. Evaluate
∫
dx√9 + x2
Example 7.3.3. Evaluate
∫ 4
2
√x2 − 4
xdx
15
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.3.4. Evaluate
∫
√4− x2
x2dx
Example 7.3.5. Evaluate
∫
dx
(5− 4x− x2)5/2
16
Chapter 7 Notes, Stewart 7e Chalmeta
7.4 Integration of Rational Functions by Partial Fractions
7.4.1 Reducing an Improper Fraction (long division)
When the degree of the numerator is greater than or equal to the degree of the denominator, we will beusing long division to simplify the integrand. Sometimes after doing long division, you will have a remainderthat will be placed on top of the divisor; this part of the new integrand may result after integration ineither a natural logarithm or an arctangent.
∫
dx
x2 + a2=
1
atan−1
(x
a
)
+ c
Example 7.4.1. Evaluate
∫
x2 + 7x− 3
x+ 4dx
Example 7.4.2. Evaluate
∫
5x5 + 28x
x4 + 9dx
17
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.4.3. Evaluate
∫ 3
−1
4x2 − 7
2x+ 3dx
7.4.2 Partial Fractions
Partial Fractions consists of decomposing a rational function into simpler component fractions and thenevaluating the integral term by term.
Example 7.4.4. Denominator is a product of disctinct linear factors∫
3x+ 7
x2 + 6x+ 5dx
18
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.4.5. Denominator is a product of linear factors, some of which are repeated.∫
3x2 − 8x+ 13
(x+ 3)(x− 1)2dx
Example 7.4.6. Denominator contains irreducible quadratic factors, none of which is repeated.∫
2x2 + x− 8
x3 + 4xdx
19
Chapter 7 Notes, Stewart 7e Chalmeta
7.4.3 Rationalizing substitutions
Some nonrational functions can be changed into rational functions by means of appropriate substitutions.
Example 7.4.7.
∫ √x
x− 4dx
7.4.4 Additional Examples
Example 7.4.8. Evaluate
∫
2x+ 1
x2 + 2x− 3dx
20
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.4.9. Evaluate
∫
x2 − x− 21
(x2 + 4)(2x− 1)dx
Example 7.4.10. Evaluate
∫
4x2 + 3x+ 6
x2(x2 + 3)dx
21
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.4.11. Evaluate
∫
ey√16− e2y
dy
Example 7.4.12. Evaluate
∫
dx
1 + ex
Example 7.4.13. Evaluate
∫
e2t
et − 2dt
22
Chapter 7 Notes, Stewart 7e Chalmeta
7.5 Strategy for Integration
7.5.1 Basic Integration Formulas
1.
∫
du = u+ C 2.
∫
k · f(x)dx = k
∫
f(x)dx
3.
∫
(du± dv) dx =
∫
du±∫
dv 4.
∫
un du =un+1
n+ 1+ C; n ∈ R, n 6= −1
5.
∫
1
udu = ln |u|+ C 6.
∫
eu dx = eu + C
7.
∫
au du =au
ln a+ C 8.
∫
sinudu = − cosu+ C
9.
∫
cosudu = sinu+ C 10.
∫
sec2 udu = tanu+ C
11.
∫
csc2 udu = − cotu+ C 12.
∫
secu tanudu = secu+ C
13.
∫
cscu cotudu = − cscu+ C
14.
∫
tanudu =
∫
sinu
cosudu = − ln | cosu|+ C = ln | secu|+ C
15.
∫
cotudu =
∫
cosu
sinudu = ln | sinu|+ C = − ln | cscu|+ C
16.
∫
secudu =
∫
secu
(
secu+ tanu
secu+ tanu
)
du =
∫(
sec2 u+ secu tanu
secu+ tanu
)
du = ln | secu+ tanu|+ C
17.
∫
cscudu =
∫
cscu
(
cscu+ cotu
cscu+ cotu
)
du =
∫(
csc2 u+ cscu cotu
cscu+ cotu
)
du = − ln | cscu+ cotu|+ C
18.
∫
du√a2 − u2
= sin−1 u
a+ C 19.
∫
du
a2 + u2=
1
atan−1 u
a+ C
20.
∫
du
u√u2 − a2
=1
asec−1
∣
∣
∣
u
a
∣
∣
∣+ C
21.
∫
du
u2 − a2=
1
2aln
∣
∣
∣
∣
u− a
u+ a
∣
∣
∣
∣
+ C 22.
∫
du√u2 ± a2
= ln∣
∣
∣u+
√
u2 ± a2∣
∣
∣+ C
24
Chapter 7 Notes, Stewart 7e Chalmeta
7.5.2 Procedures for matching integrals to basic formulas
1. Simplify the integrand.2. Make a simplifying substitution (u-substitution).3. Compete the square.4. Use a trigonometric identity / make a trigonometric substitution.5. Eliminate a square root.6. Reduce an improper fraction.7. Separate a fraction / partial fractions.8. Multiply by a form of 1.9. Use integration by parts
Example 7.5.1.
∫
xex2
dx
Example 7.5.2.
∫
xe2x dx
Example 7.5.3.
∫
x3ex2
dx
25
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.5.4.
∫
8
x2 − 6x+ 9dx
Example 7.5.5.
∫
2x2 − x+ 4
x3 + 4xdx
Example 7.5.6.
∫
dx
x2√x2 + 4
26
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.5.7.
∫
sin2 x cos5 xdx
Example 7.5.8.
∫
ex
sin(ex)dx
Example 7.5.9.
∫
dx√6− x2 + 10x
dx
27
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.5.10.
∫ π
4
0
√
1 + cos(4x)dx
Example 7.5.11.
∫
(√x+ 10)3√
xdx
Example 7.5.12.
∫
x5 sec5(3x6) tan3(3x6)dx
28
Chapter 7 Notes, Stewart 7e Chalmeta
7.6 Integral Tables
The Basic techniques of integration are substitution and integration by parts; these techniques transformunfamiliar integrals into integrals whose forms are recognizable or can be found in a table. The integraltables were created by applying substitutions and integration by parts to generic integrals in order to savethe trouble of repeating laborious calculations. When an integral matches an integral in the table or canbe changed into one of the tabulated integrals through algebra, trigonometry substitution or calculus, thetables give a ready made solution for the problem.
Example 7.6.1.
∫
dx
x√3x+ 4
Formula #
Example 7.6.2.
∫
dy√
1 + 9y2Formula #
29
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.6.3.
∫
√
5 + sin(4θ) cot(4θ)dθ Formula #
Example 7.6.4.
∫ √x√
1− xdx Formula #
30
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.6.5.
∫
dx
x3√3− x4
Formula #
Example 7.6.6.
∫
√x− x2
xdx Formula #
31
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.6.7.
∫
cot4 (3x)dx Formula #
Example 7.6.8.
∫
dx
x√7− x2
Formula #
32
Chapter 7 Notes, Stewart 7e Chalmeta
7.7 Approximate Integration
NOTE: Approximate/numerical integration is used when either we cannot find an antiderivative to aproblem or one does not exist.
7.7.1 Midpoint Rule
A. Formula∫ b
af(x)dx ≈ Mn =
n∑
i=1
f(x̄i) ·∆x = ∆x [f(x̄1) + f(x̄2) + · · ·+ f(x̄n)]
Where ∆x =b− a
nand x̄i =
xi−1 + xi
2= midpoint of [xi−1, xi].
B. The Error Estimate for the Midpoint Rule, EM .
1. EM =
∫ b
af(x)dx−Mn, where Mn is the Midpoint Rule.
2. If f ′′ is continuous and K is any upper bound for the values of |f ′′| on [a, b], then
|EM | ≤(
b− a
24
)
(∆x)2K ≤(
(b− a)3
24n2
)
K, where ∆x =b− a
n
Example 7.7.1. Use the Midpoint Rule to estimate
∫ 1
−1(x2 + 1)dx using n = 4. Find the error in the
midpoint approximation, EM .
33
Chapter 7 Notes, Stewart 7e Chalmeta
7.7.2 Trapezoidal Rule
A. Formula
One way to approximate a definite integral is by the use of n trapezoids rather than rectangles. In thedevelopment of this method we will assume that the function f(x) is continuous and positive valued on
the interval [a, b] and that
∫ b
af(x)dx represents the area of the region bounded by the graph of f(x)
and the x-axis, from x = a to x = b.
Partition the interval [a, b] into n equal subintervals, each of width ∆x =b− a
nsuch that
a = x0 < x1 < x2 < · · · < xn = b.
The area of a trapezoid is Atrap = 12h(b1+b2) where b1 and b2 are the two parallel sides of the trapezoid
and h is the distance between the two parallel sides.
b2 b1
h
With the trapezoid in this position, the height is h = ∆x and the bases b1 = f(x0) and b2 = f(x1). Wehave formed n trapezoids which have areas given by:
Area of first trapezoid: A1 =
(
1
2
)(
b− a
n
)
(f(x0) + f(x1))
Area of second trapezoid: A2 =
(
1
2
)(
b− a
n
)
(f(x1) + f(x2))
...
Area of nth trapezoid: An =
(
1
2
)(
b− a
n
)
(f(xn−1) + f(xn))
When you add up all the areas and combine like terms you get a formula that look like:
∫ b
af(x)dx ≈ Tn =
n∑
i=1
A1 +A2 + · · ·+An =∆x
2[f(x0) + 2f(x1) + 2f(x2) + · · ·+ 2f(xn−1) + f(xn)]
where ∆x =b− a
nand xi = x0 + i∆x.
Note: The coefficients in the Trapezoidal Rule follow the pattern: 1 2 2 · · · 2 1
34
Chapter 7 Notes, Stewart 7e Chalmeta
B. Error Estimate for the Trapezoidal Rule, ET .
1. ET =
∫ b
af(x)dx− Tn, where Tn is the Trapezoidal Rule.
2. If f ′′ is continuous and K is any upper bound for the values of |f ′′| on [a, b], then
|ET | ≤(
b− a
12
)
(∆x)2K ≤(
(b− a)3
12n2
)
K, where ∆x =b− a
n
Example 7.7.2. Use the Trapezoidal Rule to estimate
∫ 1
−1(x2 + 1)dx using n = 4. Find the error in the
trapezoidal approximation, ET . How large do we have to choose n so that the approximation Tn to the
integral
∫ 1
−1(x2 + 1)dx is accurate to 0.001?
35
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.7.3. Use the tabulated values of the integrand to estimate the integral
∫ 3
0
θ√16 + θ2
dθ using
the trapezoidal rule. Find the error in the trapezoidal approximation, ET .
θθ√
16 + θ2
0 0
0.375 0.09334
0.75 0.18429
1.125 0.27075
1.5 0.35112
1.875 0.42443
2.25 0.49026
2.625 0.58466
3 0.6
7.7.3 Simpson’s / Parabolic Rule
A. The formula
Another way to approximate a definite integral is by the use of n parabolas rather than rectangles ortrapezoids. In the development of this method we will assume that the function f(x) is continuous and
positive valued on the interval [a, b] and that
∫ b
af(x)dx represents the area of the region bounded by
the graph of f(x) and the x-axis, from x = a to x = b.
Partition the interval [a, b] into n equal subintervals, each of width ∆x =b− a
nsuch that
a = x0 < x1 < x2 < · · · < xn = b. For Simpson’s rule the value of n MUST BE EVEN.
36
Chapter 7 Notes, Stewart 7e Chalmeta
y0
y1
y2
h h
The area under the parabola is A = h3 (y0 + 4y1 + y2). We will use this to estimate the area under a
curve.
Simpson’s Rule (n is even)
Let f(x) be continuous on [a, b]. Simpson’s Rule for approximating
∫ b
af(x)dx is given by
∫ b
af(x)dx ≈ Sn =
(
b− a
3n
)
[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + · · ·+ 4f(xn−1) + f(xn)]
OR
∫ b
af(x)dx ≈ Sn =
(
∆x
3n
)
[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + · · ·+ 4f(xn−1) + f(xn)]
where ∆x = b−an
Note: The coefficients in Simpson’s Rule follow the pattern: 1 4 2 4 2 4 · · · 4 2 4 1
B. Error Estimate for Simpsons Rule, ES .
1. Es =
∫ b
af(x)dx− Sn, where Sn is Simpsons Rule.
2. If f (4) is continuous and K is any upper bound for the values of |f (4)| on [a, b], then
|ES | ≤(
b− a
180
)
(∆x)4K ≤(
(b− a)3
180n4
)
K, where ∆x =b− a
n
37
Chapter 7 Notes, Stewart 7e Chalmeta
Example 7.7.4. Use Simpsons Rule to estimate
∫ 1
−1(x2 + 1)dx using n = 4. Find the error in Simpson’s
approximation, ES . How large do we have to choose n so that the approximation Tn to the integral∫ 1
−1(x2 + 1)dx is accurate to 0.0001?
Example 7.7.5. Use the tabulated values of the integrand to estimate the integral
∫ 3
0
θ√16 + θ2
dθ using
Simpson’s rule. Find the error in the trapezoidal approximation, ES .
θθ√
16 + θ2
0 0
0.375 0.09334
0.75 0.18429
1.125 0.27075
1.5 0.35112
1.875 0.42443
2.25 0.49026
2.625 0.58466
3 0.6
38