Chapter 7 Notes, Stewart 7e
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Contents
7.1
7.2
Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Trigonometric Integrals
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Z
7.2.1
sinm x cosn (x) dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Evaluating
Z
tanm x secn x dx . . . . . . . . . . . . . . . . . . . . . . . . .
Z
Z
Z
7.2.3 Evaluating tan x dx,
cot x dx,
sec x dx, and
csc x dx . . . . . . .
√
√
√
Trigonometric Substitution for Integrals Involving a2 − x2 , a2 + x2 , x2 − a2
Integration of Rational Functions by Partial Fractions . . . . . . . . . . . . . . .
7.4.1 Reducing an Improper Fraction (long division) . . . . . . . . . . . . . . .
7.4.2 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4.3 Rationalizing substitutions . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4.4 Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Strategy for Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.5.1 Basic Integration Formulas . . . . . . . . . . . . . . . . . . . . . . . . . .
7.5.2 Procedures for matching integrals to basic formulas . . . . . . . . . . . . .
Integral Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Approximate Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.7.1 Midpoint Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.7.2 Trapezoidal Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.7.3 Simpson’s / Parabolic Rule . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.2
Evaluating
Z
7.3
7.4
7.5
7.6
7.7
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2
8
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Chapter 7 Notes, Stewart 7e
7.1
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Integration by Parts
Introduction
If the techniques/formulas we introduced earlier do not work then there isZanother 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
dv
du
(uv) = u
+v
dx
dx
dx
d(uv) = u dv + v du (differential form)
Z
u dv = d(uv) − v du (Solve for u dv )
Z
u dv = uv − v du (integrating both sides)
R
R
B. Note: The integration by parts expresses one integral, v du in terms of another, u dv. The idea is
that with a proper choice of u and v, the second integral is easier to integrate than the original. You
may 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
Z
Z
u dv = uv − v du
2. Definite Integrals
Z
b
b
u dv = uv
a
a
−
Z
b
v du
a
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 the
priority for u.
I -inverse trig functions
L -logarithmic functions
A -algebraic functions
T -trigonometric functions
E -exponential functions
2
Chapter 7 Notes, Stewart 7e
Example 7.1.1.
Z
3xe2x dx
Example 7.1.2.
Z
ln x dx
Example 7.1.3.
Z
e
Chalmeta
x2 ln x dx
1
3
Chapter 7 Notes, Stewart 7e
Example 7.1.4.
Example 7.1.5.
Z
Z
√
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3
tan−1 x dx
1
y 2 e3y dy
4
Chapter 7 Notes, Stewart 7e
Example 7.1.6.
Z
t4 sin (2t) dt
Example 7.1.7.
Z
ex cos (4x) dx
Example 7.1.8. Use the reduction formula
Z
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Z
n−2
1
tan u secn−2 u +
sec u du =
n−1
n−1
n
sec4 y dy
5
Z
secn−2 u du
Chapter 7 Notes, Stewart 7e
Example 7.1.9.
Z
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t3 cos (t2 ) dt
Example 7.1.10. Find the area of the region enclosed by the curve y = x cos x and the x-axis from
x = π2 to x = 3π
2
y
y = x cos(x)
x
0
−1
π
2
π
3π
2
−2
−3
6
Chapter 7 Notes, Stewart 7e
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Example 7.1.11. Find the volume of the solid generated by revolving the region in the 1st quadrant
bounded by the axes, the curve y = ex and the line x = ln 2 about the line x = ln 2.
y = ex , x = ln 2, x = 0, y = 0
2
1
x
ln 2
1
A Summary of Common Integrals Using Integration by Parts:
Z
Z
Z
n ax
n
x e dx
x sin(ax) dx
xn cos(ax) dx
Z
Z
x ln x dx
Z
x sin
eax sin(bx) dx
Z
eax cos(bx) dx
n
n
−1
(ax) dx
Z
xn tan−1 (ax) dx
7
Chapter 7 Notes, Stewart 7e
7.2
7.2.1
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Trigonometric Integrals
Evaluating
Z
sinm x cosn (x) dx
A. If the power of cosine is odd (n = 2k + 1), save one
the remaining factors in terms of sine:
Z
Z
m
2k+1
sin x cos
x dx =
Z
=
Z
=
cosine factor and use cos2 x = 1 − sin2 x to express
sinm x cos2k x cos x dx
sinm x cos2 x
k
cos x dx
sinm x 1 − sin2 x
k
cos x dx
Then substitute u = sin x.
B. If the power of sine is odd (m = 2k + 1), save one sine factor and use sin2 x = 1 − cos2 x to express the
remaining factors in terms of cosine:
Z
sin
2k+1
x cos x dx =
Z
=
Z
sin2 x
=
Z
1 − cos2 x
n
sin2k x cosn x sin x dx
k
cosn x sin x dx
k
cosn x sin x dx
Then substitute u = cos x.
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
1
[1 − cos (2x)] and cos2 x = [1 + cos (2x)]
2
2
E. It is sometimes helpful to use the identity sin x cos x =
Example 7.2.1.
Z
sin θ cos4 (θ) dθ
8
1
2
sin(2x).
Chapter 7 Notes, Stewart 7e
Example 7.2.2.
Z
sin3 φ dφ
Example 7.2.3.
Z
sin3
√
x cos3
√
x
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√
x
dx
9
Chapter 7 Notes, Stewart 7e
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Example 7.2.4. Find the area bounded by the curve of y = sin4 (3x) and the x-axis from x = 0 to x =
y
y = sin4 (3x)
π
3
1
x
π
3
7.2.2
Evaluating
Z
tanm x secn x dx
A. If the power of secant is even (n = 2k), save one factor of sec2 x and use sec2 x = 1 + tan2 x to express
the remaining factors in terms of tangent:
Z
tanm x sec2k x dx =
Z
tanm x sec2k−2 x sec2 x dx
=
Z
tanm x sec2 x
=
Z
tanm x 1 + tan2 x
Then substitute u = tan x.
10
k−1
sec2 x dx
k−1
sec2 x dx
Chapter 7 Notes, Stewart 7e
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B. If the power of tangent is odd (m = 2k + 1), save one factor of sec x tan x and use tan2 x = sec2 x − 1
to express the remaining factors in terms of secant:
Z
tan
2k+1
x sec x dx =
Z
=
Z
tan2 x
=
Z
k
sec2 x − 1 secn−1 x sec x tan x dx
n
tan2k x secn−1 x sec x tan x dx
Then substitute u = sec x.
Z
Example 7.2.5.
tan6 y dy
Example 7.2.6.
Z
π/6
tan θ sec3 θ dθ
0
11
k
secn−1 x sec x tan x dx
Chapter 7 Notes, Stewart 7e
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Example 7.2.7. Find the volume of the solid generated
by revolving the region bounded by the curve of
√
π
2
2
2
2
y = sec (x ) tan (x ), the x-axis and the line x = 2 about the y-axis.
y
y = sec2 (x2 ) tan2 (x2 )
2
1
x
√
π
2
Evaluating
Z
1. When evaluating
Z
7.2.3
tan x dx,
Z
tan x dx or
cot x dx,
Z
Z
sec x dx, and
Z
csc x dx
cot x dx rewrite the integrand in terms of sine and cosine and then
use u-substitution.
Z
sec x + tan x
, and then use
2. When evaluating
sec x dx multiply the integrand by the form of one,
sec x + tan x
u-substitution.
Z
csc x + cot x
3. When evaluating
csc x dx multiply the integrand by the form of one,
, and then use
csc x + cot x
u-substitution.
12
Chapter 7 Notes, Stewart 7e
Example 7.2.8.
Example 7.2.9.
Z
Z
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π/2
cot θ dθ
π/4
sec x dx
Integral Formulas for tan(x), cot(x), sec(x), csc(x)
Z
1.
tan u du = − ln | cos u| + C = ln | sec u| + C
2.
Z
cot u du = ln | sin u| + C = − ln | csc u| + C
3.
Z
sec u du = ln | sec u + tan u| + C
4.
Z
csc u du = − ln | csc u + cot u| + C
13
Chapter 7 Notes, Stewart 7e
7.3
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Trigonometric Substitution for Integrals Involving
√
a2 − x 2 ,
√
a2 + x 2 ,
√
x 2 − a2
Introduction
R √
Consider the integral x a2 − x2 dx. The substitution of u = a2 − x2 will allow us to integrate this
integral.
R√
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 indefinite
integral or the definite integral in which the limits do not define either a half or quarter of a circle? We
will 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
− π2 ≤ θ ≤
π
2
1 − sin2 θ = cos2 θ
a2 + x2
x = a tan θ
− π2 ≤ θ ≤
π
2
1 + tan2 θ = sec2 θ
x2 − a2
x = a sec θ
0 ≤ θ ≤ π, θ 6=
Example 7.3.1. Evaluate
R√
a2 − x2 dx
14
π
2
sec2 θ − 1 = tan2 θ
Chapter 7 Notes, Stewart 7e
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Simplifications of the trigonometric substitutions
p
√
√
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
Z
dx
√
Example 7.3.2. Evaluate
9 + x2
Example 7.3.3. Evaluate
Z
4
2
√
x2 − 4
dx
x
15
Chapter 7 Notes, Stewart 7e
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Example 7.3.4. Evaluate
Z √
Example 7.3.5. Evaluate
Z
4 − x2
dx
x2
dx
(5 − 4x − x2 )5/2
16
Chapter 7 Notes, Stewart 7e
7.4
7.4.1
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Integration of Rational Functions by Partial Fractions
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 be
using long division to simplify the integrand. Sometimes after doing long division, you will have a remainder
that will be placed on top of the divisor; this part of the new integrand may result after integration in
either a natural logarithm or an arctangent.
Z
1
dx
−1 x
+c
=
tan
x2 + a2
a
a
Z 2
x + 7x − 3
Example 7.4.1. Evaluate
dx
x+4
Example 7.4.2. Evaluate
Z
5x5 + 28x
dx
x4 + 9
17
Chapter 7 Notes, Stewart 7e
Example 7.4.3. Evaluate
7.4.2
Z
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3
−1
4x2 − 7
dx
2x + 3
Partial Fractions
Partial Fractions consists of decomposing a rational function into simpler component fractions and then
evaluating the integral term by term.
Example 7.4.4. Denominator is a product of disctinct linear factors
Z
3x + 7
dx
2
x + 6x + 5
18
Chapter 7 Notes, Stewart 7e
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Example 7.4.5. Denominator is a product of linear factors, some of which are repeated.
Z
3x2 − 8x + 13
dx
(x + 3)(x − 1)2
Example 7.4.6. Denominator contains irreducible quadratic factors, none of which is repeated.
Z
2x2 + x − 8
dx
x3 + 4x
19
Chapter 7 Notes, Stewart 7e
7.4.3
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Rationalizing substitutions
Some nonrational functions can be changed into rational functions by means of appropriate substitutions.
Z √
x
dx
Example 7.4.7.
x−4
7.4.4
Additional Examples
Z
2x + 1
Example 7.4.8. Evaluate
dx
x2 + 2x − 3
20
Chapter 7 Notes, Stewart 7e
Example 7.4.9. Evaluate
Chalmeta
x2 − x − 21
dx
(x2 + 4)(2x − 1)
Z
Example 7.4.10. Evaluate
Z
4x2 + 3x + 6
dx
x2 (x2 + 3)
21
Chapter 7 Notes, Stewart 7e
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ey
dy
16 − e2y
Example 7.4.11. Evaluate
Z
√
Example 7.4.12. Evaluate
Z
dx
1 + ex
Example 7.4.13. Evaluate
Z
e2t
dt
et − 2
22
Chapter 7 Notes, Stewart 7e
Example 7.4.14. Evaluate
Chalmeta
Z p
x2 + x + 1 dx
23
Chapter 7 Notes, Stewart 7e
7.5
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Strategy for Integration
7.5.1
Basic Integration Formulas
1.
Z
3.
Z
(du ± dv) dx =
5.
Z
1
du = ln |u| + C
u
7.
Z
au
a du =
+C
ln a
9.
Z
du = u + C
Z
du ±
Z
dv
u
cos u du = sin u + C
2.
Z
k · f (x) dx = k
Z
4.
Z
un du =
6.
Z
eu dx = eu + C
8.
Z
sin u du = − cos u + C
un+1
+ C;
n+1
10.
sec2 u du = tan u + C
12.
Z
sec u tan u du = sec u + C
11.
csc u du = − cot u + C
13.
Z
csc u cot u du = − csc u + C
14.
Z
tan u du =
Z
sin u
du = − ln | cos u| + C = ln | sec u| + C
cos u
15.
Z
cot u du =
Z
cos u
du = ln | sin u| + C = − ln | csc u| + C
sin u
16.
Z
sec u du =
Z
17.
Z
csc u du =
Z
18.
Z
du
u
√
= sin−1 + C
2
2
a
a −u
20.
Z
du
u
1
+C
= sec−1
2
2
a
a
u u −a
21.
Z
u−a
1
du
+C
=
ln
2
2
u −a
2a
u+a
sec u
sec u + tan u
sec u + tan u
csc u
csc u + cot u
csc u + cot u
n ∈ R, n 6= −1
Z
Z
2
f (x) dx
du =
Z du =
Z 19.
Z
a2
22.
Z
√
sec2 u + sec u tan u
sec u + tan u
csc2 u + csc u cot u
csc u + cot u
du = ln | sec u + tan u| + C
du = − ln | csc u + cot u| + C
du
1
u
= tan−1 + C
2
+u
a
a
√
p
du
= ln u + u2 ± a2 + C
u2 ± a 2
24
Chapter 7 Notes, Stewart 7e
7.5.2
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Procedures for matching integrals to basic formulas
1.
2.
3.
4.
5.
6.
7.
8.
9.
Simplify the integrand.
Make a simplifying substitution (u-substitution).
Compete the square.
Use a trigonometric identity / make a trigonometric substitution.
Eliminate a square root.
Reduce an improper fraction.
Separate a fraction / partial fractions.
Multiply by a form of 1.
Use integration by parts
Z
2
Example 7.5.1.
xex dx
Example 7.5.2.
Z
xe2x dx
Example 7.5.3.
Z
x3 ex dx
2
25
Chapter 7 Notes, Stewart 7e
Example 7.5.4.
Z
8
dx
x2 − 6x + 9
Example 7.5.5.
Z
2x2 − x + 4
dx
x3 + 4x
Example 7.5.6.
Z
x2
Chalmeta
dx
√
x2 + 4
26
Chapter 7 Notes, Stewart 7e
Example 7.5.7.
Z
sin2 x cos5 x dx
Example 7.5.8.
Z
ex
dx
sin(ex )
Example 7.5.9.
Z
√
Chalmeta
dx
dx
6 − x2 + 10x
27
Chapter 7 Notes, Stewart 7e
Example 7.5.10.
Z
π
4
0
p
Chalmeta
1 + cos(4x) dx
Example 7.5.11.
Z
√
( x + 10)3
√
dx
x
Example 7.5.12.
Z
x5 sec5 (3x6 ) tan3 (3x6 ) dx
28
Chapter 7 Notes, Stewart 7e
7.6
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Integral Tables
The Basic techniques of integration are substitution and integration by parts; these techniques transform
unfamiliar integrals into integrals whose forms are recognizable or can be found in a table. The integral
tables were created by applying substitutions and integration by parts to generic integrals in order to save
the trouble of repeating laborious calculations. When an integral matches an integral in the table or can
be changed into one of the tabulated integrals through algebra, trigonometry substitution or calculus, the
tables give a ready made solution for the problem.
Example 7.6.1.
Z
dx
x 3x + 4
Example 7.6.2.
Z
dy
p
1 + 9y 2
√
Formula #
Formula #
29
Chapter 7 Notes, Stewart 7e
Example 7.6.3.
Z p
5 + sin(4θ) cot(4θ) dθ
Example 7.6.4.
Z
√
Chalmeta
Formula #
√
x
dx
1−x
Formula #
30
Chapter 7 Notes, Stewart 7e
Example 7.6.5.
Z
Example 7.6.6.
Z √
Chalmeta
dx
√
3 − x4
Formula #
x − x2
dx
x
Formula #
x3
31
Chapter 7 Notes, Stewart 7e
Chalmeta
Example 7.6.7.
Z
cot4 (3x) dx
Formula #
Example 7.6.8.
Z
dx
x 7 − x2
Formula #
√
32
Chapter 7 Notes, Stewart 7e
7.7
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Approximate Integration
NOTE: Approximate/numerical integration is used when either we cannot find an antiderivative to a
problem or one does not exist.
7.7.1
Midpoint Rule
A. Formula
Z
Where ∆x =
b
a
f (x) dx ≈ Mn =
n
X
i=1
f (x̄i ) · ∆x = ∆x [f (x̄1 ) + f (x̄2 ) + · · · + f (x̄n )]
b−a
xi−1 + xi
and x̄i =
= midpoint of [xi−1 , xi ].
n
2
B. The Error Estimate for the Midpoint Rule, EM .
Z b
f (x) dx − Mn , where Mn is the Midpoint Rule.
1. EM =
a
2. If f ′′ is continuous and K is any upper bound for the values of |f ′′ | on [a, b], then
(b − a)3
b−a
b−a
2
(∆x) K ≤
K, where ∆x =
|EM | ≤
2
24
24n
n
Example 7.7.1. Use the Midpoint Rule to estimate
midpoint approximation, EM .
33
Z
1
−1
(x2 + 1) dx using n = 4. Find the error in the
Chapter 7 Notes, Stewart 7e
7.7.2
Chalmeta
Trapezoidal Rule
A. Formula
One way to approximate a definite integral is by the use of n trapezoids rather than rectangles. In the
development of this method we will assume that the function f (x) is continuous and positive valued on
Z b
f (x) dx represents the area of the region bounded by the graph of f (x)
the interval [a, b] and that
a
and the x-axis, from x = a to x = b.
Partition the interval [a, b] into n equal subintervals, each of width ∆x =
a = x0 < x1 < x2 < · · · < xn = b.
b−a
such that
n
The area of a trapezoid is Atrap = 21 h(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 ). We
have formed n trapezoids which have areas given by:
1
b−a
Area of first trapezoid: A1 =
(f (x0 ) + f (x1 ))
2
n
b−a
1
(f (x1 ) + f (x2 ))
Area of second trapezoid: A2 =
2
n
..
.
b−a
1
(f (xn−1 ) + f (xn ))
2
n
When you add up all the areas and combine like terms you get a formula that look like:
Area of nth trapezoid: An =
Z
b
a
f (x) dx ≈ Tn =
n
X
i=1
A1 + A2 + · · · + An =
∆x
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )]
2
b−a
and xi = x0 + i∆x.
n
Note: The coefficients in the Trapezoidal Rule follow the pattern: 1 2 2 · · · 2 1
where ∆x =
34
Chapter 7 Notes, Stewart 7e
Chalmeta
B. Error Estimate for the Trapezoidal Rule, ET .
Z b
f (x) dx − Tn , where Tn is the Trapezoidal Rule.
1. ET =
a
2. If
f ′′
is continuous and K is any upper bound for the values of |f ′′ | on [a, b], then
b−a
(b − a)3
b−a
|ET | ≤
(∆x)2 K ≤
K, where ∆x =
12
12n2
n
Example 7.7.2. Use the Trapezoidal Rule to estimate
Z
1
−1
(x2 + 1) dx using n = 4. Find the error in the
trapezoidal
Z 1 approximation, ET . How large do we have to choose n so that the approximation Tn to the
integral
(x2 + 1) dx is accurate to 0.001?
−1
35
Chapter 7 Notes, Stewart 7e
Chalmeta
Example 7.7.3. Use the tabulated values of the integrand to estimate the integral
the trapezoidal rule. Find the error in the trapezoidal approximation, ET .
√
θ
3
0
√
θ
dθ using
16 + θ2
θ
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
Z
Simpson’s / Parabolic Rule
A. The formula
Another way to approximate a definite integral is by the use of n parabolas rather than rectangles or
trapezoids. In the development of this method we will assume that the function f (x) is continuous and
Z b
f (x) dx represents the area of the region bounded by
positive valued on the interval [a, b] and that
a
the graph of f (x) and the x-axis, from x = a to x = b.
b−a
such that
Partition the interval [a, b] into n equal subintervals, each of width ∆x =
n
a = x0 < x1 < x2 < · · · < xn = b. For Simpson’s rule the value of n MUST BE EVEN.
36
Chapter 7 Notes, Stewart 7e
Chalmeta
y1
y2
y0
h
h
The area under the parabola is A =
curve.
h
3 (y0
+ 4y1 + y2 ). We will use this to estimate the area under a
Simpson’s Rule (n is even)
Let f (x) be continuous on [a, b]. Simpson’s Rule for approximating
Z
b
a
f (x) dx ≈ Sn =
b−a
3n
Z
b
f (x) dx is given by
a
[f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn )]
OR
Z
b
f (x) dx ≈ Sn =
a
where ∆x =
∆x
3n
[f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn )]
b−a
n
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 .
Z b
f (x) dx − Sn , where Sn is Simpsons Rule.
1. Es =
a
2. If
f (4)
is continuous and K is any upper bound for the values of |f (4) | on [a, b], then
b−a
(b − a)3
b−a
4
|ES | ≤
(∆x) K ≤
K, where ∆x =
4
180
180n
n
37
Chapter 7 Notes, Stewart 7e
Example 7.7.4. Use Simpsons Rule to estimate
Chalmeta
Z
1
(x2 + 1) dx using n = 4. Find the error in Simpson’s
−1
approximation,
ES . How large do we have to choose n so that the approximation Tn to the integral
Z 1
(x2 + 1) dx is accurate to 0.0001?
−1
Example 7.7.5. Use the tabulated values of the integrand to estimate the integral
Simpson’s rule. Find the error in the trapezoidal approximation, ES .
θ
√
θ
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
Z
3
0
√
θ
dθ using
16 + θ2