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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