5.1 Integration and estimating with Finite Sums
**In this chapter – we study a method to calculate the areas and volumes of triangles,
spheres, cones
IDEA BEHIND INTEGRATION: is that we can effectively compute many quantities
by breaking them into small pieces and then summing the
contributions from each small part.
Finite Sums – the basis for defining the integral
Area – the area of a region with a curved boundary can be approximated by summing the
areas of a collection of rectangles.
***the more rectangles used – the greater the accuracy of approximation
Methods to determine Area:
#1. 2 Rectangles - the height of each rectangle is the max value of function f, obtained
by evaluating f at the left endpoint of the subinterval [0,1]
** use b – a
n
to determine the increments in the interval
n = # of rectangles
Ex. A = 1∙½ + ¾∙½ = 7/8 = .875

this estimate is larger than true A, since the 2 rectangles contain R
UPPER SUM (Left endpoint)- is obtained by taking the height of each
rectangle as the maximum (uppermost) value of f(x) for x, a point
in the base interval of the rectangle
#2. 4 Rectangles – 4 thinner rectangles that contain the region 4 – give a more accurate
Approximation – but still larger than true A because it contains R
**1-0 = 1
4
4
Ex. A = 1∙¼ + 15/16∙¼ + ¾∙¼ + 7/16 ∙¼ = .78125 (upper sum)
#3. 4 Rectangles inside the region (R) to estimate area
Ex. function f(x) = 1-x² is decreasing – so the height of each of
These rectangles is given by the value of f at the (right endpoint)
== LOWER SUM
A≈15/16∙¼ + ¾∙¼ + 7/16∙¼ + 0∙¼ = .53125
--Therefore, the true value of A .53125<A.78125
** The lower and upper sum approximations give us estimates for the area – but also a
bound on the size of the possible error (.78125 - .53125 = .25)
#4. Another estimate to find area can be found by using the MIDPOINT RULE –
gives an estimate between the upper and lower sums
A = 63/64∙¼ + 55/64∙¼ + 39/64∙¼ + 15/64∙¼ = .671875
** Must notice if curve increasing or decreasing:
a) Increasing - use right endpoints to find upper sums and left endpoints to
find lower
b) Decreasing – use right endpoints to find lower and left to find upper
The interval [a,b] over which the function is defined is subdivided into n subintervals of
equal width (also called length) and defined by: Δx = (b – a) and f was evaluated at a
n
point in each subinterval: c1 in the first subinterval, c2 in the second subinterval, etc.
So finite sums then all take the form:
f(c1)Δx + f(c2)Δx+…+f(cn)Δx
**The table on p. 328 show that the more rectangles you use – the greater the
approximation.
----------------------------------------------------------------------------------------------Distance Traveled - if we know the velocity of a car moving down a highway
without changing direction, and want to know how far it traveled
between times t =a and t = b then distance can be found by
calculating the change in position s(b) – s(a)
1) However if don’t know antiderivative for velocity v(t), we can approximate the
Distance traveled by:
a) Subdividing the interval [a,b] into short time intervals on which the velocity
is considered to be fairly constant
b) Then approximate the distance traveled on each time subinterval with the
formula D = V x t
D = V(t1)Δt + V(t2)Δt + …+V(tn)Δt
Ex. 10 p. 333
Displacement Versus Distance Traveled – when a body changes direction 1 or more
times during the trip – use body’s speed |v(t)| to find the total distance traveled.
**velocity only gives the body’s displacement – that’s why need to use speed
Total distance traveled: |v(t1)|Δt + |v(t2)|Δt + … + |v(tn)|Δt
Average Value of Nonnegative Functions [on interval a,b]
=
1
x (Area under the graph)
b-a
5.2 Sigma Notation and Limits of Finite Sums
Sigma Notation: enables us to write a sum with many terms in the compact form:
Σ ak = a1 + a2 +…+an-1+ an
Σ = sigma = sum
k= index of summation – tells us where the sum begins and n tells us where it ends
Ex.
2
Σ
k
k+1
k=1
** lower limit of summation does not have to be 1 -- it can be any integer
Algebraic Rules for Finite Sums:
#1. Sum rule: Σ(ak + bk) = Σak + Σbk
#2. Difference Rule: Σ(ak - bk) = Σak - Σbk
#3. Constant Multiple Rule Σ cak = c∙Σak
#4. Constant Value Rule:
Σc = n∙c
Ex.
Common Summation formulas:
A) The sum of the first n integers:
for any # c
n
Σk = n(n + 1)
k=1
2
B) The first n squares:
Σk² = n(n + 1)(2n + 1)
k=1
6
C) The first n cubes:
n
Σk³ = n²(n + 1)²
k=1
4
Ex.
Limits of Finite Sums – as you noticed in the last section – the finite sum gets more
accurate as the # of terms increased and the subinterval width became thinner.
a) we calculate the limiting value as the width of the subintervals go to 0 and
their #’s approach infinity
Riemann Sums – deals with the limits of finite approximations
a) Begin with a function defined on a closed interval [a,b]
b) Subdivide the interval [a,b] into subintervals (not necessarily of = widths) and
form sums in the same way for finite approximations.
c) Denote a by x0 and b by xn so then a = x0<x1<x2<…<xn-1<xn = b
d) The set P= {x0, x1, … xn-1, xn} = Partition of [a,b]
** partition divides [a,b] into n closed subintervals
1st = [x0, x1], [x1, x2]… [xn-1, xn]
Ex.
e) then chose a point on each subinterval—ex. interval x1 pick a pt (c1)
f) each subinterval form the product f(ck)∙Δxk
n
Sp = Σ f(ck)Δxk
k=1
Riemann Sum for f on the interval [a,b] – there are many of these depending on the
Partition we choose
Norm = (||P||) to be the largest of all the subinterval widths
Regular – if every subinterval is of = width, then ||P|| is b – a
n
Ex. set P = {0, .2, .6, 1, 1.5, 2} is a partition of [0,2]
5 subintervals [0,.2], [.2,.6], [.6,1], [1,1.5], [1.5,2]
Δx1 = .2
Δx2 = .4
Δx3 = .4
Δx4 = .5
So ||P|| = .5
Δx5 = .5
Ex. Graph the function over the given interval. Partition the interval into 4 subintervals
of equal width. Then add to your sketch the rectangles associated with the Reimann
Sums given that ck is the a) left hand endpoint b) right hand endpoint c)midpoint
F(x) = x² -1 [0,2]
5.3 The Definite Integral
** Involves examining the limit of general Riemann Sums as the norm of the partition
[a,b] approaches 0
Definition of Definite Integrals – based on idea that for certain functions, as the norm of
the partitions of [a,b] approaches 0, the values of the
corresponding sums approach a limiting value, I
a) ε = represents a small positive # that specifies how close to I the Reimann sum
must be.
b) δ = a 2nd small positive # that specifies how small the norm of a partition must
be in order for that to happen
The Formal Definition of The Definite Integral as a Limit of Riemann Sums
|Σf(ck)Δxk-I| < ε
We use it this way:
1) If f is defined on the closed interval [a,b] and the limit
n
Lim
||P||->0
Σ f(c1)Δx1 exists then f is integrable on [a,b]
i=1
b
And the limit is denoted by ∫ f(x) dx
a
Limit is called the DEFINITE INTEGRAL **(is a # versus indefinite integral which is a
family of functions)
b
= upper limit
∫ f(x)dx
a = lower limit
f(x) = integrand(function) x=variable of integration
Theorem: Continuity Implies Integrability
**If a function is continuous on the closed interval [a,b] then f is integrable on [a.b]
(special cases: a discontinuous f(x) that is increasing on [a,b] and piecewise
continuous f(x))
When Not Integrable Function:
1. A function that is sufficiently discontinuous so that the region between its graph
and the x-axis can’t be approximated well by increasingly thin rectangles.
2. When the limit depends on the choice of ck
The Definite Integral as the Area of a Region:
** If f is continuous and nonnegative on the closed interval [a,b] then the area of the
region bounded by the graph of f, the x-axis, and the vertical line x=a and x=b is given
by:
b
Area= ∫f(x) dx
figure:
a
Ex. Sketch the region corresponding to each definite integral, then evaluate each integral
using a geometric formula:
3
3
a) ∫ 4 dx
2
b) ∫(x+2)dx
1
c) ∫ √4 - x² dx
0
-2
Properties/Rules of Definite Integrals
a
b
∫f(x) dx =
#1. Order of Integration
- ∫f(x) dx
b
a
3
Ex.
0
∫ (x+2) dx = 21
0
2
if
∫ (x+2) dx = -21
3
2
π
a
∫ f(x) dx = 0
#2. Zero Width Interval
ex.
a
b
∫ kf(x) dx
#3 Constant Multiple
a
b
=
k ∫f(x) dx
a
∫ sin x dx = 0
π
Or
b
b
∫-f(x) dx
-∫ f(x) dx
=
a
a
#4. Sum and Difference
b
b
∫(f(x)±g(x)) dx =
∫f(x) dx
a
a
b
c
∫f(x) dx
#5. Additivity
∫g(x) dx
a
c
+ ∫f(x) dx
a
b
±
=
∫f(x) dx
b
a
#6. Min-Max Inequality – If f has max. value f and min. value f on [a,b] then:
b
min f∙(b-a) ≤ ∫f(x) dx ≤ max f∙(b-a)
a
b
f(x) ≥ g(x) on [a,b] then
#7. Domination
b
∫f(x) dx ≥ ∫g(x) dx
a
and f(x) ≥ 0 on [a,b] then
a
b
∫f(x) dx ≥ 0
a
3
3
Ex. Evaluate ∫ (-x² + 4x – 3) dx
1
using ∫ x² dx = 26
1
3
3
3
∫ x dx = 4
∫dx = 2
1
1
Some Integrating Rules:
b
a) For f(x) = x
∫ x dx = b² - a²
a
2
2
b
b) For f(x) = x
∫ c dx = c(b-a)
where c is any constant
a
b
c) for f(x) = x²
∫ x² dx
a
= b³ - a³
3 3
where a<b
Average (Mean) Value of a Continuous Function - If f is integrable on [a,b] then its
Average value on [a,b] also called its Mean Value is
b
Av(f) = 1 ∫ f(x) dx
b-a a
Ex. find the average value of f(x) = √4 - x² 0n [-2,2]
**from before we know that A= ½πr² = ½π(2)² = 2π so
2
∫ √4 - x² dx = 2π and av(f) = 1 ∙ (2π) = ½π
-2
2-(-2)
5.4 The Fundamental Theorem of Calculus
** central theorem of integral calculus—connects integration and differentiation
using Riemann Sums.
Mean Value Theorem for Definite Integrals -- if f is continous on [a,b], then at some
point c in [a.b]. b
f(c) = 1 ∫ f(x) dx
b-a a
figure:
** value f(c) in the mean value theorem is, the average or
(mean) height of f on [a,b]
** when f≥0, the area of the rectangle is the area under the
graph from a to b
** geometrically-the theorem says that there is a #c in [a,b]
Such that the rectangle with height equal to the average
Value of the function and base width b-a has exactly the
Same area as the region beneath the graph of f from atob
** Continuous Function is the key – a discontinuous function will never equal its
average value**
Ex. find the average value of f(x) = 3x² - 2x on interval [1,4]
Fundamental Theorem of Calculus Part 1:
x
** If f is continuous on [a,b] then F(x) = ∫ f(t) dt is continuous on [a,b] and differentiable
a
on (a,b) and its derivative is f(x)
x
F′(x) = d ∫ f(t) dt = f(x)
dx a
**must have x on top of integral and # on bottom – if not need to use integral rules to get
that to happen.
Ex #31. p. 365
x
Find dy/dx if y = ∫√1 + t² dt = dy/dx=√1 + t²
a
0
√x
x
#33 y=∫sint² dt
**must be in ∫ form so by rule 2.
√x
a
-∫ sint² dt
0
Fundamental Theorem Part 2 [Evaluation Theorem] –describes how to evaluate
definite integrals.
*If f is continuous at every point of [a,b] and F is any antiderivative of f on [a,b] then:
b
∫f(x) dx = F(b) – F(a)
a
So, to calculate the definite integral of f over [a,b] by:
1. Find an antiderivative F of f
b
2. Calculate the # ∫f(x) dx = F(b) – F(a)
a
Usual notation for F(b) – F(a) is:
F(x)│ba or [F(x)]ba
Summarize in 2 ways:
x
1. d/dx∫ f(t) dt = dF/dx = f(x) – if you first integrate the function f and then differentiate
the result, you get f back again.
a
x
x
2. ∫dF/dt dt = ∫f(x) dt = F(x) – F(a) ----if you first differentiate the function F and then
a
a
Integrate the result, you get the function F back
** In a sense – the process of integration and differentiation are inverses of each other
Ex. Evaluate
√x
∫cos t dt
0
Ex #17
π/2
∫(8y² + siny) dy
-π/2
1) integrate 8y³ - cos y]π/2-π/2
Total Area - by using Antiderivatives ** if curve falls below x-axis must subtract
Ex. y = 6 –x - x²
Steps: 1) find where curve crosses x-axis so set the equation = 0 and solve
6-x-x² = 0
(3+x)(2-x) = 0
x = -3, 2 so interval is [-3,2]
2) Find area of interval
2
∫(6-x-x²) dx
-3
= 6x - x² - x³]2-3
2 3
= 205/6
** Also, note that when arch is parabola – then the area under such an arch = ⅔(bxh)
=⅔(5)(25/4) = 125/6
Finding Area Using Antiderivatives:
1. Subdivide [a,b] at the zeros of f
2. Integrate f over each subinterval
3. Add the absolute value of the integrals
Ex. x³ - 4x
-2≤x≤2
5.5 Integration by Substitution
What is the difference between a definite Integral and an Indefinite Integral?
** role of substitution in integration is comparable to the Chain Rule in differentiation
2 Methods:
#1. Pattern Recognition - Antidifferentiation of a Composite Function
* Let g be a function whose range is an interval, I, and let f be a function that is
continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I
then: ∫f(g(x))g′(x) dx = F(g(x)) + C
**If u = g(x) then du = g′(x)dx and ∫f(u) du = F(u) + C
∫f(g(x))g′(x) dx
f-outside function
function
(g(x)) = inside function
g′(x) dx = derivative of inside
** If matches the pattern, then follow the methods of indefinite antidifferentiation such
as: The Power Rule:
Ex. ∫(x²+1)²(2x) dx
Ex. ∫5cos5x dx
Recognizing the Pattern with a constant – use the constant multiple rule
∫kf(x)dx = k∫f(x) dx
Ex. Find ∫x(x² +1)² dx
**Integral is missing a factor of 2 –b/c you recognize that
g′(x) is 2x
#2 Method 2 Change of Variable – when you completely rewrite the integral in terms
of u and du u=g(x) du=g′(x)
and in the form: ∫f(g(x))g′(x)dx = ∫f(u) du = F(u) + C
Ex. Find ∫√2x – 1 dx
u=
Ex. ∫x√2x – 1 dx
Ex. ∫sin²3x cos 3x dx
Guidelines for making Changes of Variables:
du =
1. Choose a substitution u=g(x). Usually it is best to choose the inner part of a
composite function such as quantity raised to a power.
2. Compute du = g′(x) dx
3. Rewrite the integral in terms of the variable u
4. Find the results integral in terms of u
5. Replace u by g(x) to obtain an antiderivative in terms of x
6. Check your answer by differentiating
General Power Rule for Integration: If g is a differentiable function of x then:
∫(g(x))ng′(x) dx = g(x)n+1 + C
or ∫ un du = un+1 + C
n+1
n+1
Ex. ∫3(3x-1)4 dx
5.6 Substitution and area between Curves
2 methods for evaluating a definite integral by substitution:
#1. find an antiderivative using substitution, then evaluate the definite integral by
applying the Fundamental Theorem
#2. Use the process of substitution directly to definite integrals
Theorem 6: Substitution in Definite Integrals
**If g′ is continuous on the interval [a,b] and f is continuous o the range of g,
then:
b
g(b)
∫ f(g(x)) ∙ g′(x) dx = ∫ f(u) du
a
g(a)
** to use this formula – 1) make the same u substitution u=g(x) and du = g′(x) – you
would use this to evaluate the corresponding indefinite
integral.
2) then integrate the transformed integral with respect to u from
The value of g(a) (the value of u at x = a) to the value g(b)
(the value of u at x = b)
Ex. Evaluate ∫3x²√x³ + 1 dx
a) evaluate by u substitution
b) pattern recognition
Definite Integrals of Symmetric Functions:
Theorem 7: let f be continuous on the symmetric interval [-a,a] then:
a
a
a) if f is even, then ∫f(x) dx = 2∫f(x) dx
-a
(remember f(-x) = f(x))
0
a
b) if f is odd, then ∫f(x) dx = 0
(remember –f(x) = f(-x)
-a
Ex. Integral of an even function:
2
∫(x4 – 4x² + 6) dx
Even b/c f(x) = f(-x)
-2
So:
Area between curves – when we want to find the area of a region that is bounded above
By the curve y =f(x) and below by curve g(x) and on the left and
Right by the lines x=a and x=b
Definition: If f and g are continous with f(x) ≥g(x) throughout [a,b] then the area of the
Region between the curves y=f(x) and y= g(x) from a to b is the integral of
(f-g) from a to b
b
A = ∫ [f(x) – g(x)]
f(x) upper curve
a
** when applying this definition - graph the curves
a) graph reveals which curve is the upper curve f and which is lower g
Ex. find the area of the region enclosed by the lines and curve in: y = 2 and y = x²-2
Steps: 1. graph
2. Equate
x²-2 = 2
x²-4=0 then solve x = 2, x= -2
3. f(x) – g(x) = 2- (x²-2)
2
4. Integrate
∫4 - x² dx
-2
5. Evaluate
Ex. Changing the Integral to Match a Boundary Change
Y = x4 - 4x² + 4
and y = x²
Steps: 1. graph
2. set equal to each other, then set =0 and solve
3. need to then f(x) – g(x) and g(x) – f(x) because there are places where the
Curve alternates between top and bottom
4. integrate
5. evaluate
Integration with respect to y - If a regions boundary curves are described by functions of
Y, the approximating rectangles are horizontal instead of
Vertical and the basic formula has a y in place of an x and
d
A = ∫[f(y) – g(y)] dy
c
**f always denotes the right hand curve f(y)-g(y) is
Nonnegative
Ex. #54 x=y² and x = 3-2y²
1. graph
2. Equate
y² = 3-2y²
3(y²-1)=0
3(y-1)(y+1)=0
3. f(y) – g(y) = (3- 2y²) - y²
y=1, -1
1
4. Integrate
3∫1 - y² dy
-1
5. Evaluate
Combining integrals with formulas from Geometry – combine calc. and geometry.
Ex.