# MATH 151 Engineering Math I, Spring 2014 JD Kim Week14 Section 6.2 Area

```MATH 151 Engineering Math I, Spring 2014
JD Kim
Week14 Section 6.2, 6.3
Section 6.2 Area
Using Rectangles to Approximate the Area Under a Curve
Let f (x) be a function defined on the interval [a, b]. We wish to approximate
the area bounded by the curve f (x), the x-axis, x = a and x = b. We begin
by partitioning the interval [a, b] into n smaller subintervals. We call P = {a =
x0 , x1 , x2 , &middot; &middot; &middot; , xn−1 , xn = b}, where a = x0 &lt; x1 &lt; x2 &lt; &middot; &middot; &middot; &lt; xn−1 &lt; xn = b, the
partition points. For each subinterval [xi−1 , xi ], choose a representative point x∗i ,
that is x∗i is any point on the interval [xi−1 , xi ]. For each subinterval [xi−1 , xi ], we
will construct a rectangle under the curve and above the x-axis, where the height of
this rectangel is f (x∗i ) and the width is ∆xi = xi − xi−1 .
The sum of the approximating rectangles gives an approximation under the graph
n
P
f (x∗i )∆xi ,
of f (x) from x = a to x = b. Moreover, the area under the curve ≈
i=1
where n is the number of rectangles constructed. We call this sum a Riemann Sum.
1
Summary
Subdividing the interval [a, b] into n smaller subintervals
a = x0 &lt; x1 &lt; x2 &lt; &middot; &middot; &middot; &lt; xn−1 &lt; xn = b.
Then the n subintervals are
[x0 , x1 ], [x1 , x2 ], &middot; &middot; &middot; , [xn−1 , xn ].
This subdivision is called a Partition of [a, b].
∆xi = xi − xi−1
is the length of the ith subinterval [xi−1 , xi ].
The length of the longest subinterval is denoted by ||p|| and is called the norm of p,
||p|| = max{∆x1 , ∆x2 , &middot; &middot; &middot; , ∆xn }.
We define the area A as the limiting values (if it exists) of the areas of the approximating polygons,
n
X
A = lim
f (x∗i )∆xi
||p||→0
i=1
2
x∗i
Ex1) A function f , an interval, partition points, and a description of the point
within the ith subinterval are given,
1. Find ||p||.
2. Sketch the graph of f and the approximating rectangles.
3. Find the sum of the approximating rectangles.
1-1) f (x) = 16 − x2 ,
1-2) f (x) = 4 cos x,
[0, 4],
p = {0, 1, 2, 3, 4},
h πi
0,
,
2
x∗i = left endpoint.
n π π π πo
p = 0, , , ,
,
6 4 3 2
3
x∗i = right endpoint
Ex2) For the following functions set up the limit of a Riemann Sum that represents
the area under the graph of f (x) on the given interval. Do not evaluate the limit!.
2-1) f (x) = x2 + 3x − 2 on the interval [1, 4] using right endpoints.
2-2) f (x) =
√
x2 + 1 on the interval [0, 5] using right endpoints.
4
Ex3) The following limits represent the area under the graph of f (x) from x = a
to x = b. Identify f (x), a, and b.
r
n
3
3P
1+ i
3-1) lim
n→∞ n i=1
n
n
10 P
n→∞ n i=1
3-2) lim
1
3
10
1+ 7+ i
n
5
Section 6.3 The Definite Integral
If f (x) ≥ 0 on the interval [a, b], the area under the curve of f (x), above the
n
P
x-axis, from x = a to x = b is ≈
f (x∗i )∆xi . We call this sum a Riemann Sum.
i=1
Ex4) If f (x) = x + x2 , the interval [0, 4], partition points p = {0, 1, 2, 3, 4}, and
x∗i is left endpoint. Find the Riemann sum.
If f (x) ≥ 0 on the interval [a, b], then the true area under the graph of f (x) from
n
P
b−a
and x∗i is any point on
f (x∗i )∆xi , where ∆xi =
x = a to x = b is A = lim
n→∞ i=1
n
the ith subinterval. We would like to define the limit of a Riemann Sum irregardless
of whether the function is positive. To that end, we will define the definite integral.
Definition The Definite Integral
The Definite Integral of f (x) from x = a to x = b is
Z
a
b
f (x)dx = lim
x→∞
n
X
f (x∗i )∆x
i=1
b−a
and x∗i is any point on the ith subinterval. In the event f (x)
n
is positive on the interval [a, b], then the definite integral is the same as the area
bounded by f (x), the x-axis, x = a and x = b. If f (x) is not always positive on the
interval [a, b], then the definite integral is the net area.
where ∆x =
6
Remark Any Riemann Sum can approximate a definite integral. Specifically,
the Midpoint Rule can be used to approximate the definite integral.
Midpoint Rule
Z
a
where ∆x =
b
f (x)dx ≈
n
X
f (x̄i )∆x,
i=1
b−a
and x̄i is the midpoint of the ith subinterval.
n
Ex5) Use the Midpoint Rule with n = 4 to approximate
Ex6) Use Geometry to evaluate the following integrals.
R3
6-1) 0 (1 − 2x)dx
7
R5√
1
x2 + 1dx.
6-2)
6-3)
R3
−1
|x − 2|dx
R0 √
−2
4 − x2 dx
8
Theorem
1.
Z
b
c dx = c(b − a)
a
2.
Z
b
cf (x) dx = c
a
3.
Z
b
f (x) dx
a
b
a
Z
(f (x) &plusmn; g(x)) dx =
4.
Z
Z
b
a
f (x) dx &plusmn;
Z
b
g(x) dx
a
a
f (x) dx = 0
a
5.
6.
Z
Z
b
a
f (x) dx = −
b
f (x) dx =
a
Z
Z
a
f (x) dx
b
c
f (x) dx +
a
Z
b
f (x) dx
c
7. If m ≤ f (x) ≤ M for all x in the interval [a, b], then
m(b − a) ≤
Z
a
b
f (x) dx ≤ M(b − a)
9
Ex7) Find
Ex8) If
R3
1
R1√
1
x5 + x2 + 1 dx
f (x) dx = 4 and
Ex9) Write
R5
f (x) dx −
−3
R3
1
g(x) dx = −3, find
R0
f (x) dx +
−3
R6
5
Ex10) Find an upper and lower bound on
10
R3
1
(f (x) + 2g(x)) dx.
f (x) dx as a single integral.
R2√
0
x3 + 1 dx
Ex11) Express the following limits as a definite integral
n
1P
n→∞ n i=1
11-1) lim
1
2
i
1+
n
!
5
n
2i
2P
3 1+
−6
11-2) lim
n→∞ n i=1
n
11
```