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Calculus
Chapter 8
Principles of Integral Evaluation
By: Rhett Chien
Edited: Anna Levina
8.1 An Overview of Integration
methods
•
•
•
•
•
•
•
•
•
•
•
∫dx = x + C
∫xr dx = ((xr+1)/(r+1)) + C and x cannot = 1
∫ex dx = ex + C
∫r dx = r ∫ dx = rx + C
∫dx/x = ln|x| + C
∫sin x dx = -cos x + C
∫cos x dx = sin x + C
∫sec2x dx = tan x + C
∫csc2x dx = -cot x + C
∫sec x tan x = sec x + C
∫csc x cot x dx = -csc x + C
8.2 Integration by Parts
• Method of integration based on the product
•
rule for differentiation
∫f(x)g(x) dx = f(x)G(x) - ∫f’(x)G(x) dx
can also be written as:
∫u dv = uv - ∫v du
u=f(x) du=f’(x) dx
v=G(x) dv=g(x) dx
Integration by Definite Parts
b
b
b
• ∫u dv = uv| - ∫v du
a
a
a
Examples~
∫x2ex dx
u= x2
dv= ex dx
du= 2x dx v= ex
∫x2ex dx = x2ex - ∫ex 2x dx
u= 2x
dv= exdx
du= 2dx
v= ex
∫x2ex dx = x2ex - 2xex - ∫2 ex dx
= x2ex - 2xex - 2 ex + C
2
∫x2lnx dx
1
u= lnx
dv= x2 dx
du= dx/x
v= (1/3)(x)3
2
2 2
∫x2lnx dx= (1/3)(x)3lnx| - ∫ ((1/3)(x)3) ( 1/x)dx
1
1 1
2
∫x2lnx dx= (8/3)ln2 – (7/9)
1
∫Calvin dx = Hobbes - ∫Calvin dx
2 ∫Calvin dx = Hobbes
∫Calvin dx = Hobbes/2
8.3 Trigonometric Integrals
• n is + and odd
• m and n are even
∫sinnx dx
= ∫sinx*sinn-1x dx
Use sin2x=1-cos2x
Pythagorean identities
∫ sinmx cosnx dx
Use half angle formulas
• n is + and even
∫ sinnx dx
Use half angle formulas
Sin2x= (1-cos2x)/2
Cos2x= (1+cos2x)/2
• m is odd and both +
∫ sinmx cosnx dx
= ∫ sinm-1x cosnx sinx dx
sin2x=1-cos2x
U-substitution
Examples~
∫sin4x dx
= ∫ ((1-cos2x)/2)2 dx
=(1/4) ∫1-2cos2x + cos22x dx
=(1/4) ∫1-2cos2x + ((1+cos4x)/2) dx
=(1/4) (x – sin2x + (1/2)x +
(1/8)sin4x) + C
∫sin5x dx
=∫ sin4x*sinx
=∫(1- cos2x)2 sinx dx
=∫sinx*(1 - 2cos2x + cos4x) dx
Let u = cosx
du = -sinx dx
= -∫1 – 2u2 + u4 du
= -cosx + (2/3)cos3x – (1/5)cos5x
+C
8.4 Trigonometric Substitutions
Evaluating integrals containing radicals by making substitutions
involving trigonometric functions
• x=a*sinθ
x=a*tanθ
x=a*secθ
Examples~
∫dx/(x2√x2-4)
•
√x2-4 = 2 tan θ
=∫(2sec θ2tan θ) / (2 sec θ)2(2tan θ) d θ
=∫(1/4)cos θ d θ
=(1/4)sin θ + C
sin θ = (√x2-4)/x
(1/4) [(√x2-4)/x] + C
8.5 Integrating Rational Functions
by Partial Fractions
Decompose a rational function into a sum of simple rational functions that can
be integrated
Proper Rational function = P(x)/Q(x)
Rational Functions = F1(x) + F2(x) + … + Fn(x)
P(x)/Q(x)= F1(x) + F2(x) + … + Fn(x)
F1(x) + F2(x) + … + Fn(x) are in the forms of
A/(ax+b)k or (Ax+B)/(ax2 + bx + c)k
Denominators are factors of Q(x)
Sum is called the partial fraction decomposition of P(x)/Q(x) and the terms are
called partial fractions.
Determine partial fraction decomposition: determining the exact form of the
decomposition and finding the unknown constants.
8.5 continue
• Linear Factor Rule
For each factor of the form (ax+b)m the partial Fraction
decomposition contains the following sum of m partial
fractions
A1/(ax+b) + A2/(ax+b)2 + … + Am/(ax+b)m
A1 A2 Am are constants to be determined.
8.5 continue
Quadratic Factor Rule
For each factor of the form (ax2+bx+c)m the partial fraction
decomposition contains the following sum of m partial
fractions
(A1x + B1)/(ax2+bx+c) + (A2x + B2)/(ax2+bx+c)2 + … +
(Amx + Bm)/(ax2+bx+c)m
Examples~
∫(4x2 + 13x – 9)/(x3 + 2x2 - 3x) dx
=
(4x2 + 13x – 9)/(x3 + 2x2 - 3x)
= (4x2 + 13x – 9)/((x)(x+3)(x-1)
=A/x + B/(x+3) + C/(x-1)
4x2 + 13x – 9 = A(x+3)(x-1) + B(x)(x-1) + C(x)(x+3)
Let x= 0
Let x = 1
Let x = -3
A=3
C=2
B= -1
∫(4x2 + 13x – 9)/(x3 + 2x2 - 3x) dx
= ∫3/x dx - ∫1/(x+3) dx +∫2/(x-1) dx
=3ln|x| - ln|x+3| + 2ln|x-1| + C
Section 8.7 Numerical Integration;
Simpson’s Rule
Riemann Sum
b
n
∫f(x) dx = lim
∑ f(xk*)∆x
a
n->∞ k=1
Trapezoidal approximation
b
∫f(x) dx = ((b-a)/2n)(y0 + 2y1 + … + 2yn-1 + yn)
a
8.7 continue
Denoting errors of midpoint and trapezoidal approximations
b
|Em|=|∫f(x) dx - Mn| Midpoint approximation error
a
b
|Et|=|∫f(x) dx - Tn| Trapezoidal approximation error
a
8.7 continue
Theorem: Let f be continuous on [a,b], and let |Em| and |Et| be the absolute
errors that result from the midpoint and trapezoidal approximations of
b using n subintervals
∫f(x) dx
a
1.) If graph is concave up or down on (a,b) |Em| < |Et|
2.) If graph of f is concave down on (a,b), then b
Tn < ∫f(x) dx < Mn
a
3.) If graph of f is concave up on (a,b), then
b
Mn < ∫f(x) dx < Tn
a
8.7 continue
Simpson’s rule – the combination of Trapezoidal and Midpoint
approximations (best approximation for area)
S2n=(1/3)(2Mn + Tn)
=(1/3)((b-a)/2n)[yo + 4y1 + 2y2 + 4y3 + 2y4 + … + y2n]
8.7 continue
8.7 continue
Examples~
Use the Simpson Rule to find the area
2
∫e-x2 dx =???
0
Section 8.8 Improper Integrals
• Let f be a function which is continuous on the closed interval
[a, ∞). We define
If this limit exists and is finite then we say that the integral
∞
∫ f(x) dx
a
is convergent; otherwise, we say that the integral is divergent.
8.8 continue
• Let f be a function which is continuous on the closed interval
(∞, b]. We define
If this limit exists and is finite then we say that the integral
b
∫f(x) dx
-∞
is convergent; otherwise, we say that the integral is divergent
8.8 continue
Let f be a function which is continuous for all real numbers. If, for some real number c,
both of
Are convergent then we define
and we say that the integral
∞
∫ f(x) dx
-∞
is convergent; otherwise, we say that the integral is divergent.
Examples~
∞
∫dx/(x2+9) =
0
b
b
Lim
∫dx/(x2+9) = lim
(1/3)tan-1(x/3)|
b-> ∞ 0
b-> ∞
0
=.524
0
∫dx/(x-1)2 =
-∞
0
0
Lim
∫x-1)-2 dx = lim
-1/(x-1)|
b-> -∞ b
b-> -∞
b
=1
QUIZ TIMEEEEEEEE
1. ∫(x-1)/(x2-2x) dx =
a) (1/2)ln|x| + ln |x-2| + C
b) (1/2)ln|(x-2)/x| + C
e) None of these
Answer : D
2. ∫(x3)(lnx) dx =
a) (x3)(3lnx + 1) + C
b) (x4/16)(4lnx – 1) + C
e) None of these
Answer : B
c)ln|x-2| + ln|x| + C
d)(1/2)ln|(x-2)(x)| + C
c) (x4/4)(lnx – 1) + C
d) 3x2(lnx – (1/2)) + C
3. .785
∫cos2x dx=
0
a)1/2
c).643
b).393 d).893
e).143
Answer : C
4. Using the midpoint area with (n = 3) and trapezoidal area (n=6) find
the area of the function y=6x-x2
a) Midpoint= 38 Trapezoid=35
c) Midpoint= 54 Trapezoid=60
b) Midpoint= 9
Trapezoid=30
d) Midpoint= 36 Trapezoid=36
e) Midpoint= 17.5 Trapezoid=17.5
Answer: A
5. Using Simpson’s rule, find the area for the function in the previous
question
a) 38
c) 28
b) 37
d) 30
e) 112
Answer : B
6. ∫ (ex) (cosx) dx
a) (1/2)ex(cosx + cosx) + C c) ex sinx
b) (1/2)ex(sinx + cosx) + C
d) ex sinx + ex cosx
e) (1/2)ex(cosx + sinx) + C
Answer = B
7. ∞
∫dx/(1+x2) =
-∞
a) 3.142
c) 0
b) 1.571
d) .785
e) None of these
Answer : A
8. ∫1/(x2√16-x2) dx
a) (√16-x2 )/16x c) 16x/(√16-x2 )
b) 16x
d) -(√16-x2 )/16x
e) None of these
Answer : D
9. ∞
∫1/x2 dx =
1
a) 0
c) 1.5
b) 2
d) -1
e) None of these
Answer : E, real answer is 1
10. ∫ (√4-x2)/x2 dx =
a) x/( √4-x2)
c) ( √4-x2)/x)
b) sin-1(x/2)
d) -( √4-x2)/x) – sin-1(x/2) + C
e) ( √4-x2)/x) – sin-1(x/2) + C
Answer : E
Bibliography
• http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/trigintdir
•
•
•
•
ectory/TrigInt.html
http://www.math.hmc.edu/calculus/tutorials/trig_substitution/
http://ltcconline.net/greenl/courses/105/Antiderivatives/NUMINT.HT
M
http://archives.math.utk.edu/visual.calculus/4/improper.2/index.html
AP book, barons