Difference Equations to Differential Equations: Section 4.7 More on Area
Difference Equations
to
Differential Equations
Section 4.7
More on Area
In Section 4.1 we motivated the definition of the definite integral with the idea of finding
the area of a region in the plane. However, to solve the problem we restricted to a very
special type of region, namely, a region lying between the graph of a function f and an
interval on the x-axis. We will now consider the more general problem of the area of a
region lying between the graphs of two functions.
y = f( x)
y = g( x)
a
b
Figure 4.7.1 Approximating the area between y = f (x) and y = g(x)
Suppose f and g are functions defined on an interval [a, b] with g(x) ≤ f (x) for all x
in [a, b]. We suppose that f and g are integrable on [a, b], from which it follows that the
function k defined by
k(x) = f (x) − g(x)
is also integrable on [a, b]. Let R be the region lying between the graphs of f and g over
the interval [a, b] and let A be the area of R. In other words, A is the area of the region of
the plane bounded by the curves y = f (x), y = g(x), x = a, and x = b. We begin with an
approximation for A. First, we divide [a, b] into n intervals of equal length
∆x =
b−a
n
and let a = x0 < x1 < x2 < x3 < · · · < xn = b be the endpoints of these intervals. Next,
for i = 1, 2, 3, . . . , n, let Ri be the region lying between the graphs of f and g over the
1
Copyright c by Dan Sloughter 2000
2
More on Area
Section 4.7
2
1.5
1
0.5
-1.5
-1
0.5
-0.5
1
1.5
Figure 4.7.2 Region bounded by the graphs of y = 2 − x2 and y = x2
interval [xi−1 , xi ]. If Ai is the area of Ri , then
n
X
A=
Ai .
(4.7.1)
i=1
Now f (xi ) − g(xi ) is the distance between the graphs of f and g at xi , and so
(f (xi ) − g(xi ))∆x
should approximate Ai reasonably well when ∆x is small. Thus
n
X
(f (xi ) − g(xi ))∆x =
i=1
n
X
k(xi )∆x
(4.7.2)
i=1
will approximate A. Moreover, we should expect that this approximation will improve as
∆x decreases, that is, as n increases, and so we should have
A = lim
n
X
n→∞
k(xi )∆x.
(4.7.3)
i=1
But now the right-hand side of (4.7.2) is a Riemann sum, in particular, the right-hand rule
sum, and so the right-hand side of (4.7.3) converges to the definite integral of k on [a, b].
Hence we have
Z
Z
b
A=
b
(f (x) − g(x))dx.
k(x)dx =
a
(4.7.4)
a
Example Let R be the region bounded by the curves y = 2 − x2 and y = x2 , as shown
in Figure 4.7.2. Note that these curves intersect when
2 − x2 = x2 ,
Section 4.7
More on Area
3
3
2
1
R2
R1
1
2
3
4
5
-1
-2
Figure 4.7.3 Region bounded by the graphs of x = y 2 and x = y + 2
which implies that 2x2 = 2, that is, x = −1 or x = 1. Hence the two curves intersect at
(−1, 1) and (1, 1), and so we may describe R as the region between the curves y = 2 − x2
and y = x2 which lies above the interval [−1, 1]. Thus if A is the area of R, we have
Z 1
A=
((2 − x2 ) − x2 )dx
−1
Z 1
=
(2 − 2x2 )dx
−1
1
2 3
= 2x − x
3
−1
2
2
= 2−
− −2 +
3
3
8
= .
3
Example Let R be the region bounded by the curves x = y 2 and x = y + 2. These two
curves intersect when
y 2 = y + 2,
which implies that
0 = y 2 − y − 2 = (y − 2)(y + 1).
Hence the two curves intersect when y = −1 and y = 2, that is, at the points (1, −1) and
(4, 2). However, looking at Figure 4.7.3, we see that not all of R lies over the interval [1, 4].
In fact, R may √
be broken up into
√ two regions, R1 and R2 , where R1 is the region between
the curves y√= x and y = − x over the interval [0, 1] and R2 is the region between the
curves y = x and y = x − 2 over the interval [1, 4]. Thus, if A is the area of R, A1 is the
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More on Area
Section 4.7
area of R1 , and A2 is the area of R2 , then
A = A1 + A2
Z 1
Z 4
√
√
√
( x − (x − 2))dx
=
( x − (− x)) +
0
1
Z 1
=
√
2 x dx +
Z 4
√
( x − x + 2)dx
1
0
4
1 2
2 3
4 3
2
2
+
x − x + 2x
= x
2
3
3
1
0
16
2 1
4
= +
−8+8 −
− +2
3
3
3 2
9
= .
2
1
The region R in the previous example may also be described as the region lying between
the curves x = y 2 and x = y+2 over the interval [−1, 2] on the y-axis. In general, analogous
to our development above, if f and g are functions defined on an interval [c, d] on the y-axis
with g(y) ≤ f (y) for all y in [c, d], then the area A of the region bounded by x = f (y),
x = g(y), y = c, and y = d (see Figure 4.7.4), is given by
Z d
(f (y) − g(y))dy.
A=
c
d
x = g( y)
x = f ( y)
c
Figure 4.7.4 Region between the curves x = f (y) and x = g(y)
(4.7.5)
Section 4.7
More on Area
5
3
2.5
2
1.5
1
R2
R1 0.5
-1
0.5
-0.5
1
1.5
2
-0.5
Figure 4.7.5 Region bounded by the graphs of y = x3 − x and y = x2
In particular, for our previous example we have
Z 2
A=
(y + 2 − y 2 )dy
−1
2
1 3
1 2
y + 2y − y
=
2
3
−1
8
1
1
= 2+4−
−
−2+
3
2
3
9
= .
2
In this case, the second method for solving the problem is a little simpler than the first; in
general, it is often useful to look at a problem both ways and evaluate using the simpler
of the two approaches.
Example Let R be the region bounded by the curves y = x3 − x and y = x2 . These
curves intersect when
x3 − x = x2 ,
that is, when
0 = x3 − x2 − x = x(x2 − x − 1).
Hence the curves intersect when x = 0,
√
1− 5
x=
,
2
or
√
1+ 5
x=
,
2
where the latter two values were found using the quadratic formula. From the graphs in
Figure 4.7.5, we see that R may be divided into two regions, R1 and R2 , where R1 extends
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More on Area
Section 4.7
√
√
from x = 1−2 5 to x = 0 and R2 extends from x = 0 to x = 1+2 5 . Note that in R1 we
have x3 − x ≥ x2 , whereas x2 ≥ x3 − x in R2 . Thus, if A is the area of R, A1 is the area
of R1 , and A2 is the area of R2 , then
A = A1 + A2
Z 1+2√5
Z 0
3
2
(x2 − x3 + x)dx
=
√ (x − x − x )dx +
1− 5
2
=
=
1 4 1 2 1 3
x − x − x
4
2
3
0
√
0
√
1− 5
2
+
1 3 1 4 1 2
x − x − x
3
4
2
1+2 5
0
13
.
12
Problems
1. Find the area of the region bounded by the curves y = x and y = x2 .
√
1
2. Find the area of the region bounded by the curves y = x and y = x.
2
2
3. Find the area of the region bounded by the curves y = x and y = x + 2.
4. Find the area of the region bounded by the curves y = cos(x) and y = x2 .
5. Find the area of the region bounded by the curves y = sin(x) and y = x2 .
6. Find the area of one of the regions lying between the curves y = cos(x) and y = sin(x)
between two consecutive points of intersection.
7. Find the area of the region in the first quadrant bounded by y = cos(x), y = sin(x),
and x = 0.
8. Find the area of one of the regions lying between the curves y = cos2 (x) and y = sin2 (x)
between two consecutive points of intersection.
9. Let R be the region bounded by the curves y = x2 and y = 2 − x.
(a) Set up an integral to find the area of R using functions of x.
(b) Set up an integral to find the area of R using functions of y.
(c) Evaluate the simpler of the integrals in (a) and (b).
10. Find the area of the region bounded by the curves x = y 2 − 1 and x = 1 − y 2 .
11. Find the area of the region bounded by the curves x = y 2 and x = 6 − y.
12. Find the area of the region bounded by the curves y = x3 − 2x and y = x2 .
13. Find the area of the region bounded by the curves y = x4 − 4x2 and y = 3x3 .
14. To estimate the surface area of a lake, 21 measurements of the width of the lake are
made at points spaced 50 yards apart from one end of the lake to the other. Suppose
the measurements are, in order, 0, 50, 100, 120, 180, 240, 300, 250, 220, 295, 305, 265,
240, 275, 225, 180, 120, 90, 63, 40, and 0, all measured in yards. Use Simpson’s rule
to approximate the surface area of the lake.