ADAMSON UNIVERSITY

COLLEGE OF SCIENCE
MATHEMATICS & PHYSICS DEPARTMENT

Module for Math Analysis II

Area of Plane Region

This prepared module focuses on the application of definite integral, determining area of a
plane region in particular. Readers are expected to have knowledge on sketching graph
and evaluating definite integral. These pieces of knowledge are necessary to make the
application of calculus make sense.

This module provides detailed explanation and examples that are easy to comprehend.
Many problems are also presented to facilitate an active participation in the learning
process which will enable the reader to solve and practice new thoughts obtained, solve
problems confidently and master the lesson accordingly.

The reader is expected to finish this module gaining the different conceptualization and
application of integration rules in determining area of a plane region. He/ she is also
expected to clear all the problems presented in this entire module.

Hard work and perseverance…let these values play their role in your learning.
 MODULE 8: Area of Plane Region

This module presents the concept and utilization of definite integral in

determining area of a plane region.

Objectives: After reading this module, the learners should be able to:
1. Enumerate theorems involving definite integrals as an area
2. Sketch and determine the areas of regions bounded by curves in
rectangular coordinates.
3. Calculate areas of plane region on rectangular plane with correct
figures or graph

Definite Integral as Area

In our previous example on evaluating definite integrals, we have observed that it

may be positive, zero or negative like the following examples.
3 3
3
𝑥4 34 14
∫ 𝑥 𝑑𝑥 = ] = − = 20
1 4 1 4 4
𝜋
∫ cos 𝑥 𝑑𝑥 = sin 𝑥]𝜋0 = sin 𝜋 − sin 0 − 0
0
2 3 2 2
3
2𝑥 3 2(3)3 2(1)3 20
∫ (√𝑥 − 3)𝑑𝑥 = ( − 3𝑥)] = ( − 3(3)) − ( − 3(1)) = −
1 3 3 3 3
1

3
Notice the following figure and we can see that only ∫1 𝑥 3 𝑑𝑥 can be interpreted as
an area of a region.

𝑓(𝑥) = 𝑥 3 𝑜𝑛 [1, 3] 𝑓(𝑥) = cos 𝑥 𝑜𝑛 [0, 𝜋] 𝑓(𝑥) = √𝑥 − 3 𝑜𝑛 [1, 3]

Area of Plane Region Page | 1

To further establish our point, let use the figure on the left showing three regions
𝐴1 , 𝐴2 and 𝐴3 . If we are asked to solve for the
definite integral of 𝑓 from 𝑎 to 𝑏, then base on the
figure we should have
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = 𝐴1 − 𝐴2 + 𝐴3 𝑓(𝑥)
𝑎
The idea is, we are adding area with the condition
that shaded regions above 𝑥-axis are interpreted
as “positive” while the shaded regions below are
“negative”.
However if we are asked to determine the area of
bounded by 𝑓(𝑥) and 𝑥-axis on the interval [𝑎, 𝑏]
or to say the shaded regions, then base on the figure we should have
𝐴𝑟𝑒𝑎 = 𝐴1 + 𝐴2 + 𝐴3

Note: We cannot answer negative values for area problems. Either you did
wrong computation or wrong interpretation. Area is either zero or positive; 𝐴 ≥ 0.

Definite Integral as the Area of a Region

If a function 𝑓 is continuous and nonnegative on the closed interval

[𝑎, 𝑏], then the area of the region bounded by the graph of 𝑓, the 𝑥-axis, and
the vertical lines 𝑥 = 𝑎 and 𝑥 = 𝑏 isgiven by
𝑏
𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥
𝑎

Example 1:
 Determine the area bounded by 𝑓(𝑥) = 9 − 𝑥 2 ,
the 𝑥-axis, and vertical lines 𝑥 = −3 and 𝑥 = 3
as shown in the figure .
Solution: The function is clearly continuous
and nonnegative on the closed interval [−3, 3],
hence

5
𝐴𝑟𝑒𝑎 = ∫ (9 − 𝑥 2 )𝑑𝑥
−5
5
𝑥3
= (9𝑥 − )]
3 −5
= 36 sq. units

Area of Plane Region Page | 2

We may use the following guidelines in calculating area bounded by a function
𝑓(𝑥), the 𝑥-axis and the vertical lines 𝑥 = 𝑎 and 𝑥 = 𝑏 on the interval [𝑎, 𝑏].
Step 1. Determine whether the function is integrable on the given interval.
Step 2. Determine whether there are 𝑥 intercepts on the given interval excluding
𝑎, 𝑏. If none, solve for the definite integral. If there is/are, proceed to the next
step
Step 3. Determine in which subinterval/s 𝑓(𝑥) ≥ 0; the definite integral is
positive on these subintervals. Determine in which subinterval/s 𝑓(𝑥) ≤ 0,
definite integral is negative on these subintervals.
For instance, we have two 𝑥-intercepts, say 𝑐1 and 𝑐2 . Then, we have the these
subintervals: [𝑎, 𝑐1 ], [𝑐1 , 𝑐2 ], [𝑐2 , 𝑏].
Step 4. Add the absolute value of the definite integrals corresponding to a
subinterval.

Example 2:
 Determine the area bounded by 𝑓(𝑥) = 𝑥 3 − 2𝑥 2 − 𝑥 + 2, the 𝑥-axis, and
vertical lines 𝑥 = 1 and 𝑥 = 3.
Solution: The function is clearly continuous
on the given interval [1. 3], hence integrable.
By factoring/synthetic division, we’ll obtain
𝑓(𝑥) = 𝑥 3 − 2𝑥 2 − 𝑥 + 2
= (𝑥 − 1)(𝑥 + 1)(𝑥 − 3)
This implies that there is an 𝑥- intercept on the
given interval which is 2.
So we have two subintervals [1, 2] and [2, 3].
We now calculate the definite integral of 𝑓.

2 2
𝑥 4 2𝑥 3 5
∫ (𝑥 3 2
− 2𝑥 − 𝑥 + 2)𝑑𝑥 = ( − + 2𝑥)] = −
1 4 3 1
12
And;
3 3
3 2
𝑥 4 2𝑥 3 37
( )
∫ 𝑥 − 2𝑥 − 𝑥 + 2 𝑑𝑥 = ( − + 2𝑥)] =
2 4 3 2
12
Thus we have

37 5 42 7
𝐴𝑟𝑒𝑎 = | | + |− | = = sq. unit
12 12 12 2

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 Preservation of Inequality

Consider the figure on the right. If functions

𝑓 and 𝑔 are continuous on the closed interval [𝑎, 𝑏]
and 0 ≤ 𝑓(𝑥) ≤ 𝑔(𝑥) for 𝑎 ≤ 𝑥 ≤ 𝑏, then we have
the following properties.

Preservation of Inequality

i. If 𝑓 is integrable and nonnegative on the closed interval [𝑎, 𝑏], then

𝑏
0 ≤ ∫ 𝑓(𝑥)𝑑𝑥
𝑎
ii. If 𝑓 and 𝑔 are integrable on the closed interval [𝑎, 𝑏], and 𝑓(𝑥) ≤ 𝑔(𝑥) for
every 𝑥 in [𝑎, 𝑏], then
𝑏 𝑏
∫ 𝑓(𝑥)𝑑𝑥 ≤ ∫ 𝑔(𝑥)𝑑𝑥
𝑎 𝑎

 Mean Value Theorem for Integrals

This theorem simply states that among the possible inscribed and
circumscribed rectangles of the region under a curve, there is a rectangle whose
area is precise equal to the area of the region under the curve.

Mean Value Theorem for Integrals

If 𝑓 is continuous on the closed interval [𝑎, 𝑏], then there exists a number 𝑐
in the closed interval [𝑎, 𝑏] such that
𝑏
∫ 𝑓(𝑥)𝑑𝑥 = 𝑓(𝑐)(𝑏 − 𝑎)
𝑎

This theorem emphasizes the existence of the number 𝑐 not the actual value. To
further understand the theorem, refer to the following comparison.

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 Average Value
The Mean Value Theorem for Integrals leads to what we call average value
of a function 𝑓 on an interval [𝑎, 𝑏]. The mean value theorem implies that the
area under the curve on a given interval is equal to the rectangle whose height is
𝑓(𝑐) which is called the average value/
Notice that by isolating 𝑓(𝑐), we have
𝑏
1
𝑓(𝑐) = ∫ 𝑓(𝑥)𝑑𝑥
𝑏−𝑎 𝑎

Average Value of a Function on an Interval

If 𝑓 is integrable on the closed interval [𝑎, 𝑏], then the average value of 𝑓 on
the interval is
𝑏
1
∫ 𝑓(𝑥)𝑑𝑥
𝑏−𝑎 𝑎

Example 3:
 Finding average value
 Find the average value of 𝑓(𝑥) = 3𝑥 2 − 2𝑥
on the interval [1, 4].
𝑏 4
1 1
∫ 𝑓(𝑥)𝑑𝑥 = ∫ (3𝑥 2 − 2𝑥)𝑑𝑥
𝑏−𝑎 𝑎 4−1 1
1
= (𝑥 3 − 𝑥 2 ]14 )
3
1
= ((43 − 42 ) − (13 − 12 ))
3
48
= = 16 𝐴𝑛𝑠𝑤𝑒𝑟
3

Area of Plane Region Page | 5

 SAQ 1: Test yourself.
Solve the following problems. Show your complete solution.

i. Determine the area bounded by 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1, the 𝑥-axis,

and vertical lines 𝑥 = 1 and 𝑥 = 4

ii. Find the average value of 𝑓(𝑥) = 𝑥 2 + 2𝑥 + 1 on the interval

[1, 4].

Area of Region Between Two Curves

We can extend the idea of area of region under a curve to area of region
bounded by two curves.

Suppose we have two functions 𝑓 and 𝑔 that are continuous on interval [𝑎, 𝑏].
Observe the following figure. Geometrically, we can interpret the area of the
region between the two function as the area under the graph of 𝑓 subtracted by
the area under the graph of 𝑔.

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To verify, we can make use of the idea of partition as how we define definite
integrals. We can partition [𝑎, 𝑏] into 𝑛 regular subintervals of width ∆𝑥. Let us
sketch a representative rectangle of width ∆𝑥 and height 𝑓(𝑥𝑖 ) − 𝑔(𝑥𝑖 ) where 𝑥𝑖
is in the 𝑖th subinterval. The area of this
representative rectangle is

∆𝐴𝑖 = (ℎ𝑒𝑖𝑔ℎ𝑡)(𝑤𝑖𝑑𝑡ℎ) = [𝑓(𝑥𝑖 ) − 𝑔(𝑥𝑖 )]∆𝑥

Adding all 𝑛 rectangles and taking the limit as

‖𝑛‖ → 0 (𝑛 → ∞), we have
𝑛

lim ∑[𝑓(𝑥𝑖 ) − 𝑔(𝑥𝑖 )]∆𝑥

𝑛→∞
𝑖=1

Since 𝑓 and 𝑔 are continuous on [𝑎, 𝑏], 𝑓 − 𝑔 is

also continuous on [𝑎, 𝑏] so the limit exists. Thus we have
𝑛

𝐴𝑟𝑒𝑎 = lim ∑[𝑓(𝑥𝑖 ) − 𝑔(𝑥𝑖 )]∆𝑥

𝑛→∞
𝑖=1

𝑏
= ∫ [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
𝑎

The graphs 𝑓 and 𝑔 in the illustration are above 𝑥- axis. However, the same
integrand [𝑓(𝑥) − 𝑔(𝑥)] can be used as long as 𝑓 and 𝑔 are continuous and
𝑔(𝑥) ≤ 𝑓(𝑥) for all 𝑥 in the interval [𝑎, 𝑏]. This is because the height of
representative rectangle is [𝑓(𝑥) − 𝑔(𝑥)] regardless of the position of the 𝑥-axis.

Area of Plane Region Page | 7

 Area of a Region Between Two Curves

If 𝑓 and 𝑔 are continuous on [𝑎, 𝑏] and 𝑔(𝑥) ≤ 𝑓(𝑥) for all 𝑥 in [𝑎, 𝑏], then the
area of the region bounded by the graph of 𝑓 and 𝑔 and the vertical lines 𝑥 =
𝑎 and 𝑥 = 𝑏 is
𝑏
𝐴 = ∫ [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
𝑎

Example 4:
 Finding area of a region between two curves
 Find the area bounded by the graphs 𝑦 = 𝑥 2 + 2, 𝑦 = −𝑥, 𝑥 = 0 and 𝑥 =
1.
Solution: Let 𝑓(𝑥) = 𝑥 2 + 2 and 𝑔(𝑥) = −𝑥. Then 𝑔(𝑥) ≤ 𝑓(𝑥) for all 𝑥 in [0, 1] as
shown in the figure.
The area of the representative rectangle is
∆𝐴 = [𝑓(𝑥) − 𝑔(𝑥)]∆𝑥

= [𝑥 2 + 2 − (−𝑥)]∆𝑥

and the area of the region is

1
𝐴 = ∫ [𝑥 2 + 2 + 𝑥)]𝑑𝑥
0
1
𝑥3 𝑥2
= ( + 2𝑥 + )]
3 2 0
13 12 03 02
= ( + 2(1) + ) − ( + 2(0) + )
3 2 3 2
17
= sq. units
6

 Finding area of a region between two intersecting curves

We may use the following guidelines in solving problems where the
region is enclosed by the two given curves with two intersection.
Step 1. Determine the 𝑥 coordinate of the two points of intersection. These will
serve as the lower and upper limit of the definite integral for area.
Step 2. Sketch the graph to determine whether 𝑓(𝑥) ≥ 𝑔(𝑥) or 𝑓(𝑥) ≤ 𝑔(𝑥).
We can also evaluate the two function with a test value 𝑐 in between the x
coordinates in step 1; greater value implies higher graph.
Step 3. We use the formula for area between two curves.

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  Find the area bounded by the graphs 𝑓(𝑥) = 𝑥 2 + 2𝑥 and 𝑔(𝑥) = 𝑥 + 2.
Solution: The assumption in this kind of problem is that the two curves encloses
a region, otherwise there must be given vertical lines.

Step 1: Determine the 𝑥 coordinate of the two points of intersection. That is,
equate the two functions then solve for 𝑥

𝑥 2 + 2𝑥 = 𝑥 + 2
𝑥2 + 𝑥 − 2 = 0
(𝑥 + 2)(𝑥 − 1) = 0
𝑥 + 2 = 0 or 𝑥 − 1 = 0
𝑥 = −2 or 𝑥 = 1
Step 2: We evaluate the two function with a test value 𝑐 in between the 𝑥 = −2
and 𝑥 = 1. Say 𝑐 = 0 since −2 ≤ 0 ≤ 1

𝑓(0) = 02 + 2(0) = 0
𝑔(0) = 0 + 2 = 2
Hence, 𝑔(𝑥) ≥ 𝑓(𝑥)

Step 3. We use the formula for area between two curves.

1
𝐴 = ∫ [𝑔(𝑥) − 𝑓(𝑥)]𝑑𝑥
−2
1
= ∫ [(𝑥 + 2) − (𝑥 2 + 2𝑥) ]𝑑𝑥
−2
1
= ∫ [−𝑥 2 − 𝑥 + 2 ]𝑑𝑥
−2
1
𝑥3 𝑥2
= (− − + 2𝑥)]
3 2 −2
13 12 (−2)3 (−2)2
= (− − + 2(1)) − (− − + 2(−2))
3 2 3 2
7 10
= +
6 3
9
= sq. unit 𝐴𝑛𝑠𝑤𝑒𝑟.
2

Area of Plane Region Page | 9

 Finding area of a region between two intersecting curves
We may use the following guidelines in solving problems where the
region is enclosed by the two given curves with three or more intersection.
Step 1. Determine the 𝑥 coordinate of the two points of intersection. This will
give us intervals of the enclosed regions.
Step 2. Determine intervals to work on. For instance, if we have three points of
intersection whose 𝑥 coordinates are 2, 3 and 4, we’ll have intervals
[2, 3] and [3,4] corresponding to an enclosed region.
Step 3. Sketch the graph to determine whether 𝑓(𝑥) ≥ 𝑔(𝑥) or 𝑓(𝑥) ≤ 𝑔(𝑥).
We can also evaluate the two function with a test value 𝑐 in the intervals we
have in step 2; greater value implies higher graph.
Step 4. We use the formula for area between two curves on each intervals in
step 2.
Step 5. Add all the area of the bounded region.
 Find the area bounded by the graphs 𝑓(𝑥) = 3𝑥 3 − 𝑥 2 − 10𝑥 and
𝑔(𝑥) = −𝑥 2 + 2𝑥.
Solution:
Step 1. Determine the 𝑥 coordinate of the two points of intersection. That is,
equate the two functions then solve for 𝑥

3𝑥 3 − 𝑥 2 − 10𝑥 = −𝑥 2 + 2𝑥
3𝑥 3 − 12𝑥 = 0
3𝑥(𝑥 − 2)(𝑥 + 2) = 0
𝑥 = −2,0, 2

Step 2. Determine intervals to work on.

So we have intervals [−2, 0] and [0, 2] which corresponds to two bounded
region.
Step 3. We can use 𝑐 = −1 and 𝑐 = 1 as our test values for each respective
intervals.
For [−2, 0]; 𝑐 = −1, we have

𝑓(−1) = 3(−1)3 − (−1)2 − 10(−1) = 6

𝑔(−1) = −(−1)2 + 2(−1) = −3
Hence, 𝑓(𝑥) ≥ 𝑔(𝑥).
For [0, 2]; 𝑐 = 1, we have

𝑓(1) = 3(1)3 − (1)2 − 10(1) = −8

𝑔(1) = −(1)2 + 2(1) = 1
Hence, 𝑔(𝑥) ≥ 𝑓(𝑥).

Area of Plane Region Page | 10

Step 4 and 5.
0 2
𝐴 = ∫ [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥 + ∫ [𝑔(𝑥) − 𝑓(𝑥)]𝑑𝑥
−2 0
0 2
= ∫ [3𝑥 3 − 12𝑥]𝑑𝑥 + ∫ [12𝑥 − 3𝑥 3 ]𝑑𝑥
−2 0
4 0 2
3𝑥 2 2
3𝑥 4
=( − 6𝑥 )] + (6𝑥 − )]
4 −2
4 0
3(−2)4 2 2
3(2)4
= −( − 6(−2) ) + (6(2) − )
4 4
= −(12 − 24) + (24 − 12)
= 24sq. unit 𝐴𝑛𝑠𝑤𝑒𝑟.

SAQ 2: Test yourself.

Solve the following problem. Show your complete solution.

i. Determine the area bounded by 𝑓(𝑥) = 𝑥 2 + 1, 𝑔(𝑥) = −𝑥 2 , and

vertical lines 𝑥 = −2 and 𝑥 = 4

ii. Determine the area bounded by 𝑓(𝑥) = 𝑥 2 + 2𝑥, −3

and 𝑔(𝑥) = 𝑥 − 1
iii. Determine the area bounded by 𝑓(𝑥) = 3𝑥 3 + 4𝑥 2 − 6𝑥,
and 𝑔(𝑥) = −𝑥 3 + 2𝑥 2 + 6𝑥

That ends our Module 8: Area of Plane Region. We hope that you did
analyzed and internalized the discussions, and accomplished all SAQs. To
further check your learning, try answering the following assessment.

Area of Plane Region Page | 11

“Success is stumbling from failure to failure with no loss of enthusiasm.”

― Winston S. Churchill

Area of Plane Region Page | 12