REVIEW FOR MIDTERM 2
MINGFENG ZHAO
March 13, 2014
(Left, right, midpoint) Riemann sum, definite integral, net area, and numerical approximations
I. For a regular partition x0 = a < x1 < · · · < xn = b of [a, b], then
∆x =
b−a
,
n
and
xk = a + k∆x,
for all k = 0, 1, 2, · · · , n.
The Riemann sum of f (x) on [a, b] with n subintervals can be written:
f (x∗1 )∆x + f (x∗2 )∆x + · · · + f (x∗n )∆x =
n
X
f (x∗k )∆x.
k=1
For the choices x∗k ’s, we have
a. For the left Riemann sum, we have x∗k = xk−1 = a + (k − 1)∆x.
b. For the right Riemann sum, we have x∗k = xk = a + k∆x.
xk−1 + xk
a + (k − 1)∆x + a + k∆x
c. For the midpoint Riemann sum, we have x∗k =
=
= a+
2
2
1
k−
∆x.
2
Z b
II. For the definite integral
f (x) dx, we have
a
Z
(1)
b
f (x) dx := lim
a
n→∞
n
X
f (x∗k )∆x.
k=1
Z b
Remark 1. For the formula (1), it has two meanings: one is to compute
f (x) dx by using the Riemann
a
Z b
n
X1
k
f
by using the integral
f (x) dx.
sum; the other one is to compute some sum like
n
n
a
k=1
III. Let R be the region bounded by the graph of f (x) and x-axis between x = a and x = b, then
the net area of R
=
the sum of the areas of the parts of R lies above the x-axis
1
2
MINGFENG ZHAO
−the sum of the ares of the parts of R lies below x-axis on [a, b]
Z b
f (x) dx
=
a
Z
the area of R
b
|f (x)| dx.
=
a
Figure 1. Yellow=positive net area, Pink=negative net area
Z
b
f (x) dx using n subintervals:
Z b
a. The Midpoint Rule approximation to
f (x) dx using n equally spaced subintervals on [a, b] is:
IV. For the numerical computation of
a
a
M (n) := [f (m1 ) + f (m2 ) + · · · + f (mn )] ∆x =
n
X
xk−1 + xk
∆x,
f
2
k=1
xk−1 + xk
for all k = 1, · · · , n.
2
Z
b. The Trapezoid Rule approximation to
where mk =
b
f (x) dx using n equally spaced subintervals on [a, b] is:
a
"
#
n−1
X
1
1
T (n) :=
f (x0 ) +
f (xk ) + f (xn ) ∆x.
2
2
k=1
Z
c. If n is an even integer, the Simpson’s Rule approximation to
b
f (x) dx using n equally spaced
a
subintervals on [a, b] is:
S(n) := [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn )]
∆x
.
3
REVIEW FOR MIDTERM 2
3
Antiderivatives, indefinite integrals, and fundamental theorem of calculus
I. If F 0 (x) = f (x), then we say that F (x) is an antiderivative of f (x). Moreover, we have
Z
Z
f (x) dx = F (x) + C,
and
a
Z
b
b
f (x) dx = F (x)|a := F (b) − F (a).
x
II. If f (x) is continuous, then
f (t) dt is an antiderivative of f (x), and
a
Z
d
dx
!
b(x)
f (t) dt
= f (b(x)) · b0 (x) − f (a(x)) · a0 (x).
a(x)
III. Basic integrals one should remember:
Z
xp dx
 p+1

 x
+ C, if p 6= −1
p+1
=

 ln |x| + C, if p = −1.
Z
1
sin(ax) + C
a
Z
1
sin(ax) dx = − cos(ax) + C
a
Z
1
sec2 (ax) dx =
tan(ax) + C
a
Z
1
csc2 (ax) dx = − cot(ax) + C
a
Z
1 ax
eax dx =
e +C
a
Z
x
1
√
dx = sin−1
+C
a
a2 − x2
Z
1
1
−1 x
dx
=
tan
+C
x2 + a2
a
a
Z
tan(x) dx = − ln | cos(x)| + C
Z
sec(x) dx = ln | sec(x) + tan(x)| + C
cos(ax) dx
=
sec(x) tan(x) dx
=
Z
sec(x) + C.
IV. Important trigonometric identities:
sin2 (θ) + cos2 (θ) = 1,
1 + tan2 (θ) = sec2 (θ)
sin(2θ) = 2 sin(θ) cos(θ),
cos(2θ) = 2 cos2 (θ) − 1 = 1 − 2 sin2 (θ) = cos2 (θ) − sin2 (θ)
4
MINGFENG ZHAO
Substitution, trigonometric substitution, and integration by parts
I. Substitution: Let u = g(x), then du = g 0 (x) dx and
Z
Z
0
f (g(x))g (x) dx =
b
Z
f (u) du,
Z
0
and
g(b)
f (g(x))g (x) dx =
f (u) du.
a
g(a)
II. Trigonometric substitution:
a. Integrals involving a2 − x2 : Let x = a sin(θ), then dx = a cos(θ) dθ and
a2 − x2 = a2 − a2 sin2 (θ) = a2 cos2 (θ).
b. Integrals involving a2 + x2 : Let x = a tan(θ), then dx = a sec2 (θ) dθ and
a2 + x2 = a2 + a2 tan2 (θ) = a2 sec2 (θ).
c. Integrals involving x2 − a2 : Let x = a sec(θ), then d sec(θ) = sec(θ) tan(θ) dθ and
x2 − a2 = a2 sec2 (θ) − a2 = a2 tan2 (θ).
III. Integration by parts: Let u and v be differentiable, then
Z
Z
u dv = uv −
Z
v du,
and
a
b
b
u(x)v 0 (x) d = u(x)v(x)|a −
Z
b
v(x)u0 (x) dx.
a
That is,
Z
uv 0 dx = uv −
Z
vu0 dx.
where dv is the most complicated portion of the integrand that can be “ easily” integrated, and u is that portion
of the integrand whose derivative du is a “ simpler ” function than u itself. For the choice of u, please follow
the ‘ILATE’ order:
I
=
Inverse trigonometric function
L
=
Logarithmic function
A
=
Algebraic function
T
=
Trigonometric function
E
=
Exponential function
REVIEW FOR MIDTERM 2
5
Trigonometric integrals
Z
I. Evaluate
sinm (x) cosn (x) dx:
Cases
Strategy
m odd and positive, n any real number
Split off sin(x), rewrite the resulting even powers of sin(x)
in terms of cos(x), and then use u = cos(x)
n odd and positive, m any real number
Split off cos(x), rewrite the resulting even powers of cos(x)
in terms of sin(x),and then use u = sin(x)
m and n both even
Use half-angle formulas to transform the integrand into a polynomial
in cos(2x), and apply the preceding strategies once again to powers
of cos(2x) greater than 1.
Z
II. Evaluate
tanm (x) secn (x) dx:
Cases
Strategy
n even
Split off sec2 (x), rewrite the remaining even power of sec(x) in terms of tan(x),
and use u = tan(x)
m odd
Split off sec(x) tan(x), rewrite the remaining even power of tan(x) in terms of sec(x),
and use u = sec(x)
m even and n odd
Rewrite the even power of tan(x) in terms of sec(x) to produce a polynomial in sec(x),
apply reduction formula to each term
Partial fraction decomposition
p(x)
be a proper rational function in reduced form. Assume the denominator
q(x)
q(x has been factored completed over the real numbers and m is a positive integer:
After using long division, let f (x) =
• Repeated linear factor: A factor (x − r)m in the denominator requires the partial fractions
A1
A2
Am
+
+ ··· +
.
2
x − r (x − r)
(x − r)m
6
MINGFENG ZHAO
Improper integrals for infinite intervals
Definition 1 (Improper integrals for infinite intervals). Let a and b be two real numbers, then
I. If f is continuous on [a, ∞), then
∞
Z
b
Z
f (x) dx := lim
II. If f is continuous on (−∞, b], then
Z
f (x) dx.
b→∞
a
a
b
b
Z
f (x) dx.
f (x) dx := lim
a→−∞
−∞
a
III. If f is continuous on (−∞, ∞) and for any real number c, then
Z ∞
Z c
Z
f (x) dx := lim
f (x) dx + lim
a→−∞
−∞
b→∞
a
b
f (x) dx.
c
For the integrals in the above three cases, if the integral is finite, we say the integral converges; otherwise, the
integral diverges.
Definition 2 (Improper integrals for unbounded integrands). Let a and b be two real numbers, then
I. If f is continuous on (a, b] with lim f (x) = ∞ or −∞, then
x→a+
b
Z
Z
f (x) dx := lim+
f (x) dx.
c→a
a
b
c
II. If f is continuous on [a, b) with lim f (x) = ∞ or −∞, then
x→b−
Z
a
b
Z
f (x) dx := lim−
c→b
c
f (x) dx.
a
III. Let a < p < b, if f is continuous on (a, b] with lim f (x) = ∞ or −∞, then
x→p
Z
a
b
Z
f (x) dx := lim−
c→p
a
c
Z
f (x) dx + lim+
d→p
b
f (x) dx.
d
For the integrals in the above three cases, if the integral is finite, we say the integral converges; otherwise,
the integral diverges.
Z
Remark 2. For Definition 1, you can first compute
f (x) dx, then plug the upper limit and lower limit.
Z
For Case I and Case II in Definition 2, you also can first compute f (x) dx, then plug the upper limit and lower
limit.
REVIEW FOR MIDTERM 2
7
But for Case III in Definition 2, you must split the integral into two integrals, then do the same steps as before.
Differential equations and integral equations
I. Differential equations: Let’s consider
y0 =
dy
= f (x)g(y).
dx
1) If g(a) = 0 for some constant a, then y(x) ≡ a is a solution.
2) If g(y) 6= 0, then
1
dy
·
= f (x).
g(y) dx
Integrate both sides with respect to x, then
Z
Z
1
dy
·
dx = f (x) dx.
g(y) dx
By the change of variables, then
Z
1
dy =
g(y)
Z
f (x) dx.
In summary,



 1) y(x) ≡ a for some constant a such that g(a) = 0
0
Z
Z
y = f (x)g(y) =⇒
.
1


dy = f (x) dx.
 2)
g(y)
II. For an integral equation, you will get a differential equation after differentiating the equation, and then solve
this differential equation.
8
MINGFENG ZHAO
Something you should remember
• Midpoint Rule approximation:
M (n) := [f (m1 ) + f (m2 ) + · · · + f (mn )] ∆x,
xk−1 + xk
for all k = 1, · · · , n.
2
• The Trapezoid Rule approximation:
where mk =
"
#
n−1
X
1
1
T (n) :=
f (x0 ) +
f (xk ) + f (xn ) ∆x.
2
2
k=1
• If n is an even integer, the Simpson’s Rule approximation:
S(n) := [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 4f (xn−1 ) + f (xn )]
∆x
.
3
• Derivative of the integrals:
d
dx
Z
!
b(x)
f (t) dt
= f (b(x)) · b0 (x) − f (a(x)) · a0 (x).
a(x)
• Basic integrals one should remember:
Z
 p+1

 x
+ C,
p
p+1
x dx =

 ln |x| + C,
Z
if p 6= −1
if p = −1.
1
sin(ax) + C,
a
Z
1
sec2 (ax) dx = tan(ax) + C,
a
Z
x
1
√
dx = sin−1
+ C,
a
a2 − x2
Z
tan(x) dx = − ln | cos(x)| + C,
Z
sec(x) tan(x) dx = sec(x) + C.
cos(ax) dx =
Z
eax dx =
1 ax
e +C
a
Z
1
sin(ax) dx = − cos(ax) + C
a
Z
1
csc2 (ax) dx = − cot(ax) + C
a
Z
x
1
1
dx = tan−1
+C
2
2
x +a
a
a
Z
sec(x) dx = ln | sec(x) + tan(x)| + C
• Important trigonometric identities:
sin2 (θ) + cos2 (θ) = 1,
1 + tan2 (θ) = sec2 (θ)
sin(2θ) = 2 sin(θ) cos(θ),
cos(2θ) = 2 cos2 (θ) − 1 = 1 − 2 sin2 (θ) = cos2 (θ) − sin2 (θ)
REVIEW FOR MIDTERM 2
9
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca