6. Integration
6.3-4 Riemann Sums, Estimating Distance Traveled, and the Definite Integral
6.3 Estimating Distance Traveled
Example 1. Getting acquainted: Suppose a car travels at a constant velocity of 65 miles per hour
for 4 hours. How far does the car travel in 4 hours?
Notice that the distance traveled by the car is equal to the area between the curve y = 65 and the
x-axis on [0, 4]. Thus, the distance traveled is equal to the area between the curve y = v(t) and the
t-axis on the interval [a, b].
Example 2. Suppose an object travels with a velocity in feet per second given by v = 1t on the t interval
[1, 4] where t is in seconds.
1.Estimate the distance traveled taking left hand sums for n = 6. Sketch the rectangles associated with
this sum. Round to four decimal places.
2.Estimate the distance traveled taking right hand sums for n = 6. Sketch the rectangles associated with
this sum. Round to four decimal places.
1
Riemann Sums
WHY
Want to find the area of the region that lies between the curve y = f (x) (with curved sides) and the
x−axis between x = a and x = b.
Example 3. Find the area under the curve y = x2 on [0,1].
HOW
We will approximate the area of the region by breaking the region into polygons (in particular, rectangles), then taking limit of the sum of areas of the polygons.
We now approximate the area under the parabola y = x2 , where x belongs to interval [0,1] as in example
1.
Step 1: Divide the interval [0,1] into 5 subintervals of equal length (step size).
Step 2: Consider the rectangles whose bases are these subintervals and whose heights are the values
of the function at the right-hand endpoints of these subintervals.
2
Step 3: Find the area of each rectangle.
Step 4: Find R5 , the sum of the areas of the rectangles. This sum is an approximation of the area
under the parabola.
Right Hand Sum (RHS): Rn (with n the number of used rectangles). Use the right endpoint rk of
each subinterval Ik to determine the heights of the rectangles.
Left Hand Sum (LHS): Rn (with n the number of used rectangles). Use the left endpoint lk of each
subinterval Ik to determine the heights of the rectangles.
LHS and RHS are special cases of Riemann sums. A Riemann sum is found by using an x-value
in each subinterval to determine the height of each rectangle.
The larger n (i.e. the more rectangles), the better approximation. Thus, we define the area to be the
limit of the sums of the areas of the approximating rectangles when n → ∞.
NOTE: It is not really necessary that we use rectangles of equal width, just that the number of rectangles
increases to infinity.
Example
R 104. Use a left sum and right sum with recgangle of equal width for n = 5 to approximate the
integral 1 (2x2 + 1) dx, round the answers to two decimal places.
3
6.4 The Definite Integral
We saw that the area between the curve y = f (x) and the x-axis is given by the limit of the Riemann
sums as the number of rectangles increases. It turns out that this same type of limit occurs in a wide
variety of situations, even when f is not necessarily a positive function.
Given a continuous function f , the definite integral of f on the interval [a, b] is
Z
b
f (x) dx = lim
a
n→∞
n
X
f (xk )∆xk
k=1
where a and b are the lower limit and upper limit of the integration, respectively. The interval [a, b]
is called the interval of integration.
Rb
Note: a f (x) dx is all one expression, gives a scalar value, not depend on x. In fact, we could use any
letter in place of x without changing the value of the integral.
An integral need not represent an area. But for a non-negative function, an integral can be interpreted
as an area. That is, if f (x) ≥ 0, then the area of the region between the graph of y = f (x) and the
x-axis on the interval [a, b] is
Z b
A=
f (x) dx.
a
In general, a definite integral can be interpreted as a difference of areas:
A1 − A2 ,
where A1 is the area of the region above the x-axis and below the graph of f and A2 is the area of the
region below the x-axis and above the graph of f .
Example 5. Write a definite integral to represent the area in example 1.
4
Example 6. Find the value of
Z
3
p
( (x) + x)dx,
1
by finding the area of the appropriate geometric region.
Properties of the Definite Integral :
• If f (x) ≥ 0 for a ≤ x ≤ b, then
Rb
f (x) dx ≥ 0.
Rb
Rb
• If f (x) ≥ g(x) for a ≤ x ≤ b, then a f (x) dx ≥ a g(x) dx.
a
• If m ≤ f (x) ≤ M for a ≤ x ≤ b, then
Z
m(b − a) ≤
b
f (x) dx ≤ M (b − a).
a
Example 7. Use the fact that
evalue each of the integrals.
R 3√
a) 1 π dx
b)
Rb
x dx =
a
Rb
b 3 − a3
b 2 − a2
and a x2 dx =
and properties of integral to
2
3
R3 2
(x + 3x − 1) dx
1
5
c)
R3
1
f (x) dx, where
(
−x + 3
f (x) =
2x2
if 1 ≤ x < 2
if 2 ≤ x ≤ 3.
Example 8. Write the given sum or difference as a single integral in the form
a)
Z
3
Z
f (x) dx +
1
5
Z
f (x) dx +
3
7
f (x) dx
5
b)
Z
7
Z
f (x) dx −
3
f (x) dx
1
6
4
Rb
a
f (x) dx.
Example 9. Use the properties of integrals to compare
integrals.
7
R3
1
x2 dx and
R31
dx, without evaluating the
1
x