Spring 2016 Math 152
Overview of Material for Test I
courtesy: Amy Austin
1. Integration by Substitution: (Section 6.5) If
u = g(x) is a differentiable function, then
Z
f (g(x))g ′ (x) dx =
Z
2. Area: (Section 7.1)
•A=
b
[T op − Bottom] dx if an x-integral is pre-
a
ferred and the curves do not intersect between a
and b. If the curves intersect at x = c for a < c < b,
then the integral must be split at x = c.
• A =
Z
(a) V =
d
[Right − Lef t] dy if a y-integral is pre-
c
ferred and the curves do not intersect between c and
d. If the curves intersect at y = e for c < e < d,
then the integral must be split at y = e.
3. Volume: (Section 7.2-7.3)
Z
b
2πxf (x) dx if revolving around the y
a
axis.
(b) V =
Z
d
2πyg(y) dy if revolving around the x
c
axis.
f (u) du.
Typically, u is chosen so that du is a factor of the
integrand.
Z
• Shell:
(c) If revolving around a line (other than the x or
y axis), adjust the radius of the cyndrical shell
accordingly.
• Slicing: The base of the solid will be described, as
well as a cross section A of the solid. You will
Z inteb
grate within the boundary of the base. V =
A dx
a
if cross sections are perpendicular to the x axis, and
V =
Z
d
A dy if cross sections are perpendicular to
c
the y axis.
4. Work: (Section 7.4) If the force F is constant, then
the work W done in moving the object a distance d
is W = F d. If the force f (x) is not constant, then
the work W doneZ in moving the object from x = a
b
f (x) dx.
to x = b is W =
a
•Disk:
(a) V =
Z
b
π(f (x))2 dx if revolving around the x
a
axis.
(b) V =
Z
• Spring problems: The force f (x) needed to
maintain a spring stretched x units beyond its
natural length is f (x) = kx. Therefore the work W
required to stretch the spring from x =Za to x = b
b
d
2
π(g(y)) dy if revolving around the y
a
c
axis.
(c) If revolving around a line (other than the x or
y axis), adjust the radius of the circular disk
accordingly.
• Washer:
(a) V =
Z
b
π((f (x))2 − (g(x))2 ) dx, where
a
f (x) ≥ g(x) for a ≤ x ≤ b and revolving
around the x axis.
(b) V =
Z
d
π(g(y))2 − (h(y))2 ) dy where
c
g(y) ≥ h(y) for c ≤ y ≤ d and revolving
around the y axis.
(c) If revolving around a line (other than the x
or y axis), adjust the radius of the outer and
inner circle accordingly.
f (x) dx.
units beyond the natural length is W =
• Rope pulling problems: Depends on what is
being done in the context of the problem. To be as
general as possible: If the rope is b units long and
weighs l (Pounds or Newtons) per unit of length
with an object of weight W attached (note: if no
weight is attached, then W = 0), then the work
required to pull entire rope and weight is
W =
Z
b
((b − y)l + W ) dy =
0
Z
b
(bl + W − yl) dy.
0
• Water pumping problems:
W = ρg
Z
dA dy, where ρg is the weight density
of water, A is the area of an arbitrary (horizontal)
slice of water and d is the distance the slice of water
must travel to reach the top of the tank.
5. Average Value: (Section 7.5)
• The average value of f (x) from x = a to x = b is
Z b
1
f (x) dx.
fave =
b−a a
• The Mean Value Theorem for Integrals states that
if f (x) is continuous over the interval [a, b], then
there is a number c, a ≤ c ≤ b so that f (c) = fave .
6. Integration by Parts: (Section 8.1)
Formula:
Z
u dv = uv −
Z
v du. You much choose
7. Trig Integrals (Section 8.2)
• Integrals of the form
Z
sinm x cosn x dx:
a.) If m is odd (and positive), factor out one sine
and use sin2 x = 1 − cos2 x. Then, substitute
u = cos x.
b.) If n is odd (and positive), factor out one cosine
and use cos2 x = 1 − sin2 x. Then, substitute
u = sin x.
u and dv.
Helpful acronym: LIPET. In order, let u equal the
word that follows. Then dv is rest of the acronym.
d.) If both m and n are even, use the identities:
L=logarithm
I=Inverse trig function
P=Polynomial
• Case I:
(b)
(c)
Z
Z
xn ekx dx: u = xn ; dv = ekx dx
xn sin(kx) dx: u = xn ; dv = sin(kx)dx
xn cos(kx) dx: u = xn ; dv = cos(kx)dx
• Case II:
Z
xn (ln x) dx: u = ln x ; dv = xn dx
• Case III:
(a)
(b)
(c)
Z
Z
Z
arccos x dx: u = arccos x ; dv = dx
arcsin x dx: u = arcsin x ; dv = dx
arctan x dx: u = arctan x ; dv = dx
• Case IV: The Loop! This requires two iterations
of integration by parts.
Z
ex cos x dx or
Z
cos2 x =
Z
1
(1 + cos 2x)
2
tanm x secn x dx:
a.) If m is odd (and positive), factor out one
sec x tan x and use tan2 x = sec2 x − 1. Then, substitute u = sec x.
T=Trigonometric
(a)
1
sin2 x = (1 − cos 2x)
2
• Integrals of the form
E=Exponential (ie ex )
Z
c.) If both m and n are odd, use either case above
(but not both).
ex sin x dx, choose u however you
wish. When you integrate by parts again, choose
u as you did in the first iteration of integration by
parts.
b.) If n is even (and positive), factor out one sec2 x
and use the identity sec2 x = 1 + tan2 x. Then,
substitute u = tan x.
c.) If m is odd and n is even, use either case above
(but not both).
d.) If m is even and n is odd, try breaking up into
sines and cosines.