Name
MAT 182
Integration Notes
In Calculus I, you should have explored the notion of area. For example, say you want to compute
the area of the shaded region R, pictured below, that lies between the x-axis, the graph of y = 1 – x2
and the y-axis.
There is no simple geometric formula (like there is for the area of a circle, for example) that will
compute this area. However, you can approximate this region using rectangles of arbitrarily small
width and heights lying on the function. For example, below is a picture of 16 left-hand rectangles
used to approximate the area of R.
As the number of rectangles approaches infinity, the area of the rectangles approaches the actual area
of the region, i.e.
n 1
b
i 0
a
lim  f ( xi )x   f (x ) dx .
n
See examples of computing areas using Riemann Sums at: http://www.youtube.com/watch?v=2qhs_03OcUs&feature=youtu.be and
http://www.youtube.com/watch?v=Zzn2YcxIW6A&feature=youtu.be
The Definite Integral
The area between a function and the x-axis on the closed interval [a, b] is
represented by the definite integral,
b
 f ( x ) dx .
a
To evaluate the integral means to find the given area. The lower bound a is called
the lower limit of integration, while b is the upper limit of integration. The
function is called the integrand. The variable accompanying d is called the variable
of integration, in this case x.
b
Note that

a
b
b
a
a
f ( x ) dx   f (t ) dt   f (u) du .
Page 1 of 41
Theorem:
A continuous function is integrable. That is, if a function f is continuous on an
interval [a, b], then its definite integral over [a, b] exists.
The converse of this theorem is not necessarily true. There are discontinuous functions which are
integrable and there are discontinuous functions which are NOT integrable.
Properties of Definite Integrals
b

1. Order of Integration:
a
a
f ( x ) dx    f ( x ) dx
b
a
 f (x ) dx  0
2. Zero Width Interval:
a
b
b
a
a
 k  f (x ) dx  k  f (x ) dx
3. Constant Multiple:
b
b
b
  f (x )  g(x) dx   f (x) dx   g(x ) dx
4. Sum and Difference:
a
a
a
b
c
c
a
b
a
 f (x ) dx   f (x ) dx   f (x ) dx
5. Additivity:
Finally, we note one final property of definite integrals. A definite integral of a function can be
represented as the “signed area” of the region bounded by its graph. Thus, regions above the x-axis
are positive while regions below the x-axis are negative.
b
Thus, for the picture below,
 f (x) dx  A
1
 A2 , but the area of the region is A1 + A2.
a
Page 2 of 41
The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects integration and differentiation, and allows us to
compute definite integrals by using antiderivatives.
The Fundamental Theorem of Calculus
If a function f is continuous on the closed interval [a, b] and F is an antiderivative of
f on the interval [a, b], then
b
 f (x) dx  F(b)  F( a) .
a
To see the proof of this theorem, see http://www.youtube.com/watch?v=yhrohL5a4Fo
For each of the following, let k be a constant and let C be the constant of integration.
Function
General Antiderivative
f (x)  k
F(x)  kx  C
1 n1
f ( x )  x n , n  1
F( x ) 
x C
n1
1
F(x)  ln x  C ; x  0
f ( x )  x 1 
x
f (x)  sin x
F(x)   cos x  C
f (x)  cos x
F(x)  sin x  C
f (x)  ex
F( x )  e x  C
f ( x )  sec 2 x
F(x)  tan x  C
f (x)  sec x tan x
1
f (x) 
1  x2
1
f (x) 
1  x2
F(x)  sec x  C
F( x )  sin 1 x  C
F( x )  tan 1 x  C
1. Use the Fundamental Theorem of Calculus (and the table of common antiderivatives) to
evaluate the following definite integrals.
1
a.
  1  x  dx
2
4
b.
1
0

c.
2
  sec   d
2
4
0
  3 x  dx
d.
  2t
2
 3t  2  dt
1
Page 3 of 41
0
e.
 x
1
3
x
2
3
1
2
 dx
f.
   2  t  t  dt
0
4
1
u2
du
g. 
u
1
h.
x x
dx
3
0

***You are now ready for Homework #1: Section 6.4 # 10, 14, 16, 20, 30
The Second Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus
x
If f is continuous on [a, b] then G( x )   f (t ) dt is continuous on [a, b] and differentiable
a
on (a, b) and its derivative is f(x):
G '( x ) 
x

d 
  f (t ) dt   f ( x ) .
dx  a

Let’s look at the first part of this theorem. G(x) is a function on x, where x is the upper bound of a
definite integral. This is the first time we are seeing a function defined this way. This means that the
function G(x) gives the “signed area” under the graph of f from t = a to t = x as pictured below:
y
y = f (t)
signed area
= G(x)
a
x
b
t
The Second Fundamental Theorem of Calculus states that the derivative of this “area” function is f,
itself.
Page 4 of 41
2. Use the Second Fundamental Theorem of Calculus to find the following.
x
x


d 
d 
1
dt 
a.
b.
  cos t dt 

2
dx  
dx  0 1  t


5

d 
c.
  3t sin t dt 
dx  x

x

d 
*d.
  cos t dt 

dx  1

4

d 
1
dt
e.
 

dx  1 3 x2 2  et 
x

d 
f.
  t 2  5t dt 

dx   x2

2
3
To see more examples, watch http://www.youtube.com/watch?v=PGmVvIglZx8
***You are now ready for Homework #2: Section 6.4 # 54, 72, 76, 78, 80
Indefinite Integrals and the Substitution Rule
Recall that the set of all antiderivatives of f is the indefinite integral of f with respect to x denoted,
2
 f (x ) dx . For example,  2x dx  x  C .
Also recall the chain rule for differentiation:
d
 f g ( x)  f ' g ( x)  g ' ( x)
dx
Page 5 of 41
We will now learn a technique called integration by substitution which will allow us to “undo” the
chain rule.
We begin with an example:  3x 2 cosx 3 dx .
We can see that the above looks like the result of applying the chain rule. The “inside” function (or g)
is x3 so g(x) = x3 and g’(x) = 3x2, which appears as a factor in the above integrand. Also the “outside”
function (or f) is cosine, whose antiderivative is sine, so the above integral is exactly g’(x)·f’(g(x)).
Therefore the general antiderivative is f(g(x)) which is sin (x3) + C.
Notice that we may check the result by taking the derivative to get:
d
sin x 3  C  cos x 3  3 x 2
dx
So we know that  3 x 2 cosx 3 dx  sin x 3   C
   
 
Once you notice that an integrand is the result of chain rule, you may use a method of integration
called substitution. For the example above, substitution would work as follows:
Let u = x3 (the inside function), then
du
 3x 2 which means that du = 3x2 dx. So the integral becomes:
dx
2
3
 3x cos  x  dx =  cos(u) du which we can easily integrate as follows:
= sin(u) + C . Finally, substitute back in terms of x to get:
= sin(x3) + C.
Example 2: Find the indefinite integral:
x
2
cos( x 3 ) dx
In the example above, we can do the same substitution, but notice that we are “missing” a “3” inside
the integral, we can fix this as follows:
1
1
1
1
2
3
2
3
3
 x cos(x ) dx  3  3x cos(x ) dx  3  cos(u) du  3 sin(u)  C  3 sin(x )  C .
3. Integrate the following.
a.
 te
t 1dt
c.

 4x
2
1  2x
2
dx
b.
 xx
d.
 cos
2
2
 1 dx
2
x sin x dx
Page 6 of 41
e.
3
2
 5x 1  x dx
x2
g.
 1  x 
i.
 cos
3 2
dx
sin x
dx
2
x
3
f.
 1  1 
 1  t   t 2 dt
h.
 tan
j.
 x
2
4
x  sec 2 x dx
x 1
dx
 2 x  3
Page 7 of 41
NOTICE: The following integrals can be solved without substitution:
x 2  3x  7
 2
k. 
dx
l.  t 2  t   dt
 t
x
4. Integrate the following.
a.
  2 x  1
c.
ex
 e x  4 dx
2
dx
b.
x
d.
 6t t
1  x dx
2
3
 1 dt
19
Page 8 of 41
e.
g. 
x
 xe dx
2
1
1
cos  dt
2
t
t 
x2
f.

h.
t  2t 2
 t dt
x 2  4x  7
dx
To use substitution, we might first need to simplify an expression, or use algebra to make substitution
possible.
5. Integrate the following.
dx
ln x 2
a.  x  x
b. 
dx
e e
x
Page 9 of 41
We may also need to take advantage of trigonometric identities to transform integrals into a form that
allows substitution.
6. Integrate the following.
a.  sin 2 ( x ) dx
b.  cos 2 ( x ) dx
dx
c.
 cos (2 x )
d.
 tan
e.
 sin (x) dx
f.
 tan x dx
2
3
2
( x ) dx
For more examples, go to http://www.youtube.com/user/rootmath/featured and search for “substitution”
***You are now ready for Homework #3: Section 6.5 # 12, 16, 20, 32; Section 8.3 # 8, 19, 36;
Section 11.3 # 6, 10, 20, 24
Page 10 of 41
Integration by Parts
The method of substitution reverses chain rule. Now we need a method that reverses product rule.
d
u  v   u ' v  uv'
Recall the product rule:
dx
Rewriting this formula and taking the antiderivative gives us the following:
Integration by Parts
 uv' dx  uv   u ' v dx
7. Confirm the formula for integration by parts.
 xe
Example 1:
x
dx
Notice that neither algebra nor substitution can help us evaluate this indefinite integral. (Convince
yourself that this is true.)
This integral is a product. One of our terms, x or ex, must be u and the other v’.
We’d like to choose u and v’ so that the integral of ( u 'v ) is something we CAN integrate. Think
about it. Should we let u = x or ex ? Why?
Now, let u = x, v’ = ex so that
u'= 1, v = ex + C
x
So,  xe dx  xe x  C    1  e x  C  dx  xe x  C   e x  Cx   C1  xex  Cx  e x  Cx  C1  xex  e x  C1
Notice that the original C’s cancel, and so we could have let v be just ex. From now on, we will omit
the C in v.
2
Example 2:
 xe dx
x
0
When dealing with a definite integral, we need to evaluate the entire antiderivative at the endpoints,
so with the above we know that
2
 xe dx  xe
0
x
x
 e x    2 e 2  e 2    0  e 0   e 2  1
2
0
Page 11 of 41
8. Integrate the following.
a.
x
2
ln x dx
b.

ln x
dx
x2
1
c.
 x cos x dx
1
dx
e. 
3
xln x 
et
d.  2 dt
t
f.

ln x 2 dx
x
Page 12 of 41
*g.
i.
 ln x dx
h.
2
 ln x  dx
9. Use integration by parts twice to integrate
j.
x
2
 x sec
2
x dx
xe2 x
 2 x  12 dx
sin x dx .
Page 13 of 41
10. Integrate  x 3 e 2 x dx .
If many repetitions of integration by parts are required, the calculations can be cumbersome. In
situations like this, there is a way to organize the calculations, called tabular integration by parts.
Example:  x 3 e x dx
Alternate Signs
+
+
u and its derivatives
x3
3x2
6x
-
6
+
0
-
0
v' and its antiderivatives
ex
ex
ex
ex
ex
ex
We continue the columns until the derivative is zero.
Following the arrows,  x 3 e x dx  x 3 e x  3x 2 e x  6 xe x  6e x (+ 0 + 0 +…) + C. This table method is a
compact way of doing integration by parts 5 times!
11. Explain why the first column has alternating signs.
Page 14 of 41
12. Use the “table” method to integrate
x
3
cos x dx .
There is another special case of integration by parts. We explore this case next.
13. a. Use integration by parts twice for the following. (Don’t use a table.)
 e sin x dx
x
b. Once you’ve done this, you should notice a relationship between the original integral and your
result. Use this relationship to algebraically solve for  e x sin x dx .
Page 15 of 41
14. Use the same technique to integrate  e 3 x cos( 2 x) dx
In some cases, we can integrate a single term using integration by parts. For example,
 ln( x) dx  x ln x  x , as we saw earlier by letting u = ln x, and v’ = 1.
15. Use the same idea to integrate the following.
a.  sin(ln x) dx
b.
 sin
1
x dx
To see more examples, go to http://www.youtube.com/user/bullcleo1?feature=watch or http://www.youtube.com/user/patrickJMT/featured and
search for “integration by parts”
***You are now ready for Homework #4: Section 8.2 # 12, 20, 28, 34, 36, 62
Page 16 of 41
Trigonometric Substitution
3
16. We can evaluate

9  x 2 dx exactly, using geometry. What is it?
3
17. However, we cannot find

9  x 2 dx . Why?
18. Try the substitution x = 3 sin() so that dx = 3 cos() d in the integral from # 16. (Recall that
+ cos2(x) = 1, so cos2(x) = 1 – sin2(x).)
sin2(x)
Page 17 of 41
In fact, in general, integrals involving a2  u2 where a is a positive constant and u is a function of x,
we can draw the following right triangle:
u
, so multiplying
a
both sides by a gives us a sin  = u. We
make this our substitution.
Notice that sin  
a
u

a2  u2
Since u = a sin , then du = cos  d and so the expression in our integrand becomes:




a2  u2  a2  ( a sin )2  a2  a2 sin 2   a2 1  sin 2   a2 cos2   a cos  . (This can be found
a 2  u2
.)
a
19. Integrate the following using a similar trigonometric substitution.
directly from our triangle by noting that cos  
x
1
2
25  x 2
dx
Page 18 of 41
We can use trigonometric substitutions to evaluate integrals involving
a2  u2 ,
a2  u2 and
u2  a2 . We summarize the substitutions below.
1.
For integrals involving
u = a sin(), then
du = cos() d and
a2  u2 , let
a2  u2  a cos() .
a
u

a2  u2
2. For integrals involving
u = a tan(), then
du = sec2() d and
a2  u2 , let
a 2  u2  a sec() .
3. For integrals involving
u = a sec(), then
a2  u2

u2  a2 , let
du = sec()tan() d and u  a   a tan() ,
with the positive result if u > a and the
negative result if u < -a.
2
2
a
u

u
u2  a2
a
20. Integrate the following using an appropriate substitution.
dx
dx
a. 
b. 
3
(4  x 2 )2
x2  1


Page 19 of 41
c.
e.

x
3x 3
1x
2
dx
dx
2
x2  1
d.
f.
x

1  x 2 dx
x2  4
dx
x
To see more examples, go to http://www.youtube.com/user/patrickJMT/featured and search for “trigonometric substitution”
***You are now ready for Homework #5: Section 8.4 # 22, 24, 30, 34; Hint for #30: Once you do a
trigonometric substitution, convert the integrand to sines and cosines (instead of tangents and secants) then you will
need a substitution.
Page 20 of 41
Integration Techniques applied to Definite Integrals
Using the Fundamental Theorem of Calculus and substitution, we can now evaluate more
complicated definite integrals.
Substitution in Definite Integrals
If g’ is continuous on the interval [a, b] and f is continuous on the range of g, then
b

f  g( x )   g '( x ) dx 
a
g( b )

f (u) du .
g( a )
21. Evaluate the following definite integrals.
ln 2
1
a.
2
3
 3x x  1 dx
b.
1
e
3x
dx
0
The integration by parts formula can be combined with the Fundamental Theorem of Calculus in
order to evaluate definite integrals.
Integration by Parts Formula for Definite Integrals
b

a
b
f ( x ) g '( x ) dx   f ( x ) g( x )a   f '( x ) g( x ) dx
b
a
22. Find the area of the region bounded by the curve y  xe  x and the x-axis on the interval [0,4].
Page 21 of 41
Let f be continuous on the symmetric interval [-a, a].
a

(a) If f is even, then
a
a
f ( x ) dx  2  f ( x ) dx . (see Figure (a) below.)
0
a
(b) If f is odd, then
 f (x) dx  0 .
(see Figure (b) below.)
a
2
23. Evaluate
x
4
 4 x 2  6 dx .
2
Area Between Curves
Suppose we want to find the area of a region that is bounded above by the curve y = f(x), below by
the curve y = g(x), and on the left and right by the lines x = a and x = b.
b
We know that the area under the curve y= f(x) on [a, b] is given by
 f (x) dx .
But we don’t want all of
a
that area. We want to “erase” the area under the curve g(x) on [a, b]. Thus, the shaded region above
b
must be given by:
  f (x)  g(x) dx .
a
Page 22 of 41
If the two curves intersect twice, you can also use this idea to find the area between the two curves.
24. Sketch and find the area of the region enclosed by the parabola y = 2 – x2 and the line y = -x.
25. Sketch and find the area of the region in the first quadrant that is bounded above by y  x and
below by the x-axis and the line y = x – 2.
26. Find the region described in the previous exercise by integrating with respect to y.
Page 23 of 41
27. Find the area of the region enclosed by the curves x  12 y 2  12 y 3 and x  2 y 2  2 y .
28. Find the area of the region enclosed by x + y = 2 and y = x2.
Page 24 of 41
29. Find the area of the region enclosed by y  x 2  4 and y   x 2  2 x and the vertical line x = -3.
30. Find the area of the region enclosed by the curves y  x 2  3x and y  2 x 3  x 2  5x .
See more examples at http://www.youtube.com/watch?v=bHvcN9XLHCQ and http://www.youtube.com/watch?v=6d5I9udgbyg&feature=youtu.be
***You are now ready for Homework #6: Section 7.1 # 2, 6, 17
Page 25 of 41
Volumes by Slicing
Now, we define volumes of solids whose cross-sections are plane regions. A cross-section of a solid
S is the plane region formed by intersecting S with a plane.
Suppose we want to find the volume of a solid like the one above. Recall from geometry that the
volume of a cylinder is given by the area of its base times its height.
Volume = base area × height = A · h
This equation forms the basis for defining the volumes of many solids that are not cylindrical, by the
method of slicing. If the cross-section of the solid S at each point x in the interval [a, b] is a region
R(x) of area A(x), and A is a continuous function of x, we can define and calculate the volume of the
solid S as a definite integral:
Volume
The volume of a solid of known integrable cross-sectional area A(x) from x = a to x =
b is the integral of A from a to b,
b
V   A(x ) dx .
a
To Calculate the Volume of a Solid with a Given Cross-Section:
1. Sketch the solid and a typical cross-section.
2. Find a formula for A(x), the area of a generic cross-section.
3. Find the limits of integration.
4. Integrate A(x) using the Fundamental Theorem of Calculus.
See http://www.youtube.com/watch?v=4lxS5slaFVQ&feature=youtu.be for more.
Page 26 of 41
31. A pyramid 3 meters high has a square base that is 3 meters on a side. The cross-section of the
pyramid perpendicular to the altitude x meters down from the vertex is a square x meters on a
side. Find the volume of the pyramid. (Check your answer with the familiar geometric
1
formula V  b 2 h .)
3
Typical cross-section
x
x
3
3
3
32. Find the volume of the solid whose base is bounded by the graphs of y = x + 1 and y  x 2  1
a. with square cross-sections
b. with cross-sections that are rectangles of height 1
Page 27 of 41
33. Find the volume of the solid whose base is the region bound by the functions y  x 2 and
y  8  x 2 whose square cross-sections are perpendicular to the x-axis.
34. Find the volume of the solid whose base is the region bound by the curves x = y2 and x = 4
a. whose circular cross-sections are parallel to the y-axis.
b. whose cross-sections are squares with bases in the xy-plane.
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35. Find the volume of the solid whose base is a unit circle, and whose cross-sections are
a. squares whose diagonal are in the xy-plane.
b. isosceles right triangles
c. squares with bases in the xy-plane
See more examples at http://youtu.be/R1zd1_jHCoM, http://youtu.be/TUvMmA5qwss, and http://youtu.be/VC37kWAK_pQ and also at
http://www.youtube.com/watch?v=DTDSZmPH7_E&feature=youtu.be and http://www.youtube.com/watch?v=puSlVA6mwNQ
***You are now ready for Homework #7: Section 7.2 # 72
Page 29 of 41
Volume by Rotation About an Axis
If a region in the x-y plane is revolved about a line, the resulting solid is a solid of revolution, and the
line is called the axis of revolution. Many familiar solids can be formed in this manner:
We can find the volumes of these solids of revolution by slicing the region. For this purpose, observe
that the cross section of any solid taken perpendicular to the x-axis at the point x is a circular disk of
radius f(x). The area of the region is A(x)
= [f(x)]2. Thus, the volume of the solid is
b
V    [ f ( x)]2 dx
a
Because the cross sections are diskshaped, the application of this formula is
called the method of disks or the disk
method.
DISK METHOD:
To Calculate the Volume of a Solid of Revolution:
1. Sketch the solid and a typical cross-section (circular).
2. Find a formula for R(x), the radius of a generic cross-section.
3. Find the limits of integration.
2
4. Integrate   R( x ) using the Fundamental Theorem of Calculus.
EXAMPLE 1: Find the volume of the solid that is obtained when the region under the curve y  x
over the interval [1,4] is revolved about the x-axis.
Solution:
b
4
4
V    [ f ( x)] dx    ( x ) dx    x dx 
2
a
2
1
1
 x2 
4
 15
 8  

2 1
2
2
Page 30 of 41
Not all solids of revolution have solid interiors; some have
holes that create interior surfaces. We can solve this problem
again by slicing. This time, however, we observe that the cross
section of the solid taken perpendicular to the x-axis at the
point x is a “washer-shaped” region with inner radius g(x) and
outer radius f(x). Hence, its area is
A(x) = [f(x)]2 - [g(x)]2 = ([f(x)]2 - [g(x)]2)
Following the previous example, what is the volume of this
solid?
V=
See http://youtu.be/R6hD2jpr19g for another visual.
WASHER METHOD:
To Calculate the Volume of a Solid of Revolution with a hole:
1. Sketch the solid and a typical cross-section (washer).
2. Find formulas for R(x), the large radius of a generic cross-section and r(x), the small radius
of a generic cross-section.
3. Find the limits of integration.


4. Integrate   R(x)  r(x) 2 using the Fundamental Theorem of Calculus.
2
EXAMPLE 2: Find the volume of the solid generated when the region between the graphs of the
equations f(x) = ½ x + x2 and g(x) = x over the [0,2] is revolved about the x-axis.
Solution:
First graph the region.
When revolved about the x-axis, will the cross-sections be discs or washers? Explain.
So we see that solution will simply be:
2
42
2
5
4
3
V    ([ x  x ]  x )dx    x 4  x 3  34x dx    x5  x4  x4  
0
5
0
0
2
1
2
2 2
2
2


Page 31 of 41
EXAMPLE 3: The axis of revolution need not always be the coordinate axis. Find, for example, the
volume of the solid formed by revolving the region bounded by f ( x)  2  x 2 and g ( x)  1 about the
line y = 1.
Solution:
By setting f(x) = g(x), we can determine the points of intersection. Do that in the space below.
Then sketch the region.
Notice that the radius of this solid is f(x) – g(x).
b
1
So, V     R( x)  dx    1  x
2
a
1

2 2
1

2 x3 x5 
16
dx    1  2 x  x  dx    x 
  
3
5  1 15

1
1
2
4
36. Find the following volumes.
a. The region between the curve y  x and the x-axis on the interval [0, 4] when revolved about the
x-axis.
b. The solid generated by revolving the region bounded by y  x and the lines y = 1 and x = 4
about the line y = 1.
Page 32 of 41
c. The solid generated by revolving the region between the parabola x= y2 + 1 and the line x = 3 about
the line x = 3.
d. The region bounded by the curve y = x2 + 1 and the line y = -x + 3 when revolved about the x-axis.
e. The region bounded by the parabola y = x2 and the line y = 2x when revolved about the y-axis.
See more examples at http://www.youtube.com/watch?v=lMYvmeeF0GM and http://www.youtube.com/watch?v=LdDyJ3X-isU&feature=youtu.be
and http://www.youtube.com/watch?v=nZqOKc067Z8
***You are now ready for Homework #8: Section 7.2 #2, 6, 10, 12c, 12d
Page 33 of 41
Arc Length
Definite integrals are used to find the length of a curve. A rectifiable curve is one that has a finite arc
length. Any function whose derivative is continuous on an interval [a, b] will be rectifiable. (There
are rectifiable curves whose derivative is not continuous.) Such a function is continuously
differentiable and its graph on that interval is smooth.
Let the function given by y = f(x) represent a smooth curve on the interval [a, b]. The arc length of f
between a and b is
b
s   1   f '( x ) dx .
2
a
Similarly, for a smooth curve given by x = g(y), the arc length of g between c and d is
d
s   1   g '( y ) dy .
2
c
For more, see http://www.youtube.com/watch?v=seoFxrNL85c.
3
37. Find the arc length of the curve f (x)  x 2 from (1, 1) to (2, 2 2 ) .
3
38. Find the arc length of the curve g( y )  13 ( y 2  2) 2 from y =0 to y = 1.
Page 34 of 41
39. Find the arc length of the graph of y  ln(cos(x)) from x= 0 to x 
of sec(x) is ln(sec(x)  tan(x)) .)

. (Hint: An antiderivative
4
For more examples, see http://www.youtube.com/watch?v=PwmCZAWeRNE.
Surface Area
We used integration to calculate the volume of a solid of revolution. Now we look at finding the
surface area of a solid of revolution.
Let y = f(x) have a continuous derivative on the interval [a, b]. The area S of the surface of
revolution formed by revolving the graph of f about a horizontal or vertical axis is
b
S  2  r( x ) 1   f '( x ) dx
2
a
where r(x) is the distance between the graph of f and the axis of revolution.
interval [c, d], then the surface area is
d
If x = g(y) on the
S  2  r( y ) 1   g '( y ) dy
2
c
where r(y) is the distance between the graph of g and the axis of revolution.
40. Find the area of the surface formed by revolving the graph of f(x) = x3 on the interval [0,1]
about the x-axis.
Page 35 of 41
41. Find the area of the surface of formed by revolving the graph of f(x) = x2 on the interval
0, 2  about the y-axis.


For more, see http://www.youtube.com/watch?v=-j2eKo84Ef8.
***You are now ready for Homework #9: Section 7.4 # 4, 6, 38
Improper Integrals
There are two types of improper integrals:
 a definite integral whose lower and/or upper bound is infinite,
 a definite integral whose integrand has an infinite range with the bounds of integration.
Examples:
1

1
ln x
(a)  2 dx
(b) 
dx
x
x
1
0
We cannot use the Fundamental Theorem of Calculus because the functions are not continuous on
the given interval or the bounds are not real numbers. However, we can use limits to get around this
problem. Let’s look at the example above.
Page 36 of 41
 b ln x 
ln x
dx

lim
dx  , as illustrated below:

1 x 2
b   x 2
1


Example (a): Notice that
b
Use integration by parts with u = ln x, v’ =
1/x2
ln x
dx . Then take the limit as b
x2
1

to evaluate
approaches infinity.
1
Example (b): Notice that the function y 
is not continuous at x = 0, and thus
x
1
improper integral. However, just as above,

0
1
Since

a
1

0
1
dx is an
x
 1

1
dx  lim  
dx  , as illustrated below:
a 0
x
a x 
1
1
1
dx   2 x   2  2 a , we can evaluate the area in terms of
a
x
1
the lower bound a. Now, to evaluate



0
1
dx , we simply need to
x
evaluate lim 2  2 a  2  2 0  2 .
a 0
1
Thus,

0
1
dx  2 .
x
Page 37 of 41
How to evaluate improper integrals:
Step 1: Identify what makes the integral improper (i.e., an infinite bound, or a discontinuity on the
interval).
Step 2: Rewrite the integral so that it is proper using a and/or b to replace the “improper bound”.
(This may require breaking the integral into two parts.)
Step 3: Evaluate the proper integral in terms of a and/or b.
Step 4: Take the limit as a and/or b approaches the original bound.
42. Evaluate the following improper integrals.
dx
a.  2
x
1


b.
dx
x
1


c.
x
  1  x  e dx
0
0
d.
ex
 3  2 ex dx
Page 38 of 41

e.



g.

0
x
e
x
3
2
x
x


2
dx
f.
2


dx
dx
 1 x
h.
x
2
1
dx
ln x
Page 39 of 41
43. Evaluate the following improper integrals.
dx
a. 
1x
1
4
2
2
c.
b.
1

dx
  x  1
d.
2
1

0
6

0
0
e.
dx
 x  2
2
3
cos x
dx
1  2 sin x
dx
1  x2
For more, see http://www.youtube.com/watch?v=85-HNJyuyrU
***You are now ready for Homework #10: Section 8.8 #12, 14, 20, 24, 30
WARNING!!!!
It is sometimes tempting to apply the Fundamental Theorem of Calculus directly to an improper
integral without taking the appropriate limits. To illustrate what can go wrong, we ignore the fact
that the following integral is improper and incorrectly evaluate the integral as:
2
2
dx NO! 
1 
0 (x  1)2   x  1  0  1  1  2
Page 40 of 41
This result is clearly incorrect because the integrand is never negative and hence the integral cannot
be negative! Hopefully you evaluated this integral correctly in part c above.
44. Use the formula for arc length to derive the formula for the circumference of a circle of radius
r. (Hint: The equation of a circle centered at (0,0) with radius r is given by x2 + y2 = r2.)
45. Let R be the region to the right of x = 1 that is bounded by the x-axis and the curve y 
1
.
x
When this region is rotated about the x-axis, it
generates a solid whose surface is known as Gabriel’s
Horn, shown to the left. The volume of this region
(which can be found using the disk method) is given

2
1
by: V      . The surface area is given by
x
1

2
1
 1 
SA  2 
1    2  dx , which is infinite.
x
 x 
1
Despite the fact that Gabriel’s Horn has an infinite
surface area, show that it has a finite volume.
2

1
 1 
1    2  dx is infinite is to notice that for x >1,
(Note: The way to show that 2 
x
 x 
1

2


1
1
1
 1 
2 
1    2  dx  2 
1 dx  2  dx , which we showed diverged in #44b.)
x
x
x
 x 
1
1
1
Page 41 of 41