Applications of Integration: Area between curvesmachiang/1014/2019_20/MATH1014...Area between curves...
Transcript of Applications of Integration: Area between curvesmachiang/1014/2019_20/MATH1014...Area between curves...
Applications of Integration:Area between curves
(Mainly based on Stewart: Chapter 6, §6.1)
Edmund ChiangMATH1014
September 4, 2019
1 Area between two curves
1.1 Area as defined by Riemann sums
Recall that how we define the area between x = a and x = b under a given function y = f(x):
Figure 1: (Stewart: Illustration of an irregular shape area S)
We also recall
Definition. Let [a, b] to denote the closed interval: a ≤ x ≤ b. Apartition of [a, b] is a set of arbitrary points as denoted by
P = {x0, x1, x2, . . . , xn−1, xn}, ∆xi = xi−xi−1 =b− an
, i = 1, · · · , n.
In particular, we have a = x1 ≤ x2 ≤ · · · ≤ xn−1 ≤ xn = b.
We again recall the definition of a (upper/lower) Riemann sum of f over the interval
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[a, b]
Definition: Let f(x) be a continuous function defined on [a, b].Then we define a Riemann sum of f(x) over [a, b] with respectto partition P to be the sum
AP = f(x∗1) ∆x1 + f(x∗2) ∆x2 + · · ·+ f(x∗n) ∆xn
where x∗i is an arbitrary point lying in [xi−1, xi], i = 1, 2, · · · , n.Depending on the choices of x∗i , we have (i)Lower Riemann sum if f(x∗) equals
minxi−1≤x≤xi
f(x)
for each i; (ii) Upper Riemann sum if f(x∗) equals
maxxi−1≤x≤xi
f(x)
for each i.
Definition: If f is a function defined on [a, b]. We divide the[a, b] into n subintervals with equal width ∆x = b−a
n . Let
P : x0 = A < x1 < x2 < · · · < xn = b
be a partition of [a, b]. Let x∗j (j = 1, · · · , n) be the sam-ple points in the subinterval [xj−1, xj]. Then we define thethe definite integral of f from a to b by∫ b
a
f(x) dx = limn→∞
n∑j=1
f(x∗j)∆x,
provided that this limit exits and it equals to the same valuefor all possible choices of sample points x∗j . When this is so, wesay that f is integrable on [a, b].
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The above definition is equivalent to∫ b
a
f(x) dx = lim∆x→0
n∑j=1
f(x∗j)∆x,
where ∆ = max{xk − xk−1}.
We have
Theorem: If f is continuous or has at most a finite numberof jump discontinuities over [a, b], then f is integrable on [a, b],and so its value, called the definite integral∫ b
a
f(x) dx
exists.
2 Area between curves
2.1 Via vertical strips
One can easily extend the above idea to find area between two curves :
Figure 2: (Stewart: An illustration of an area between f(x) and g(x))
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According to the meaning of Riemann sum, the limit
A = limn→∞
∞∑j=1
[f(x∗j)− g(x∗j)
]∆x
when exists represent the area between y = f(x) and y = g(x) over a ≤ x ≤ b. So we define
Definition: The area A of the region bounded by the curvesy = f(x) and y = g(x) over a ≤ x ≤ b, where both f and g arecontinuous functions such that f(x) ≥ g(x) for all x in [a, b] is
A =
∫ b
a
[f(x)− g(x)
]dx.
Remark . We note that if g(x) = 0 for all x in [a, b], then the above A simply representsthe area under y = f(x) and between x = a and x = b.
Figure 3: (Stewart: An illustration of an area between f(x) and g(x))
Example . Find the area of the region bounded above by y = ex, bounded below by y = x,and bounded on the sides by x = 0 and x = 1. We first sketch the graphs of y = ex andy = x.
Since the two curves intersect at x = 0 and x = 1. So a = 0 and b = 1. Then theFundamental Theorem of Calculus implies that the area is given by
A =
∫ 1
0
(ex − x) dx = ex − 1
2x2∣∣∣10
= e− 1
2− 1
= e− 3
2.
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Figure 4: (Stewart: An illustration of an area between y = ex and y = x)
Example . Find the area of the region enclosed by y = x2 and y = 2x−2.
Figure 5: (Stewart: An illustration of an area between y = 2x− x2 and y = x2)
We first find the intersecting points by solving the two equations. Hence
x2 = 2x− x2.
That is equivalent to solving 2x(x− 1) = 0. Hence x = 0 (y = 0) or x = 1 (y = 1). A sample”rectangular area” is given by[
f(x∗)− g(x∗)]∆x = (2x− 2x2) ∆x.
Hence the total area of the region is given by∫ 1
0
(2x− 2x2) dx = x2 − 2
3x3∣∣∣10
= 1− 2
3=
1
3.
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Example . Find the area of the region enclosed by y =x√x2 + 1
and y = x4 − x.
It turns out that the equation
x√x2 + 1
= x4 − x
can not be solved exactly. An numerical method gives x = 0 and x ≈ 1.18. Thus anapproximate area of the region is given by∫ 1.18
0
x√x2 + 1
− (x4 − x) dx ≈ 1
2
∫ 2.39
1
du√u−∫ 1.18
0
(x4 − x) dx ≈ 0.785.
Figure 6: (Mathematica plot: An illustration of the area between y = x√x2+1
and y = x4−x)
Example . Find the area of the region bounded by the curves y = sinx, y = cosx, x = 0and x = π/2.
We notice that cosx > sinx when 0 ≤ x < π/4, and cos x < sinx when π/4 < x ≤ π/2and cos x = sinx at x = π/4. Hence
A =
∫ π/2
0
| cosx− sinx| dx
=
∫ π/4
0
cosx− sinx dx+
∫ π/2
π/4
sinx− cosx dx
= [sinx+ cosx]∣∣∣π/40
+ [− cosx− sinx]∣∣∣π/2π/4
= 2√
2− 2.
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Figure 7: (Stewart: An illustration of the area between y = sinx and y = cosx)
Exercise . Find the area of the region bounded by the curves y = x, y = 5x− x2.We first solve the two equations. Hence
5x− x2 = x.
Thus, the solutions are given by solving x(x− 4) = 0. Thus
A =
∫ 4
0
(5x− x2)− x, dx
=
∫ 4
0
(4x− x2), dx
= 2x2 − 1
3x3∣∣∣40
=32
3.
Figure 8: (Mathematica plot: An illustration of the area between y = 5x− x2 and y = x)
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2.2 Via horizontal strips
Suppose we are given two curves x = f(y) and x = g(y). Then
Figure 9: (An illustration of an horizontal strip between two curves)
Hence the area of the region bounded by the two curves is given by
A =
∫ y2
y1
f(y)− g(y), dy.
Example . Find the area enclosed by y = x− 1 and y2 = 2x+ 6.We first work out the coordinates where the two curves intersect: x = y + 1 and x =
12(y2 − 6). That is, we want to solve y + 1 = 1
2(y2 − 6) or 0 = y2 − 2y − 8 = (y − 4)(y + 2).
The coordinates are (−1, −2) and (5, 4).
Figure 10: (An illustration of an horizontal strip between two curves)
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Hence
A =
∫ y2
y1
(x2 − x1) dy
=
∫ 4
−2(y + 1)− 1
2(y2 − 6) dy
=
∫ 4
−2(−1
2y2 + y + 4) dy = 18.
Let us use vertical strips for finding the area. There are two types of vertical stripsindicated by pale green and dark green colours in the figure. We have
• −3 ≤ x ≤ −1. The upper and lower y−coordinates are given by{upper
√2x+ 6
lower −√
2x+ 6
Thus, the area stems from type I (pale green) rectanges
AI =
∫ −1−3
√2x+ 6− (−
√2x+ 6) dx = 2
∫ −1−3
√2x+ 6 dx
Similarly, the type II (dark green) rectangles has
• −1 ≤ x ≤ 5. The upper and lower y−coordinates are given by{upper
√2x+ 6
lower x
Hence
AII =
∫ 5
−1(√
2x+ 6− x) dx.
It can be verified (as exercise) that the total area A = AI + AII = 18.
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