Integral Calculus
APPLICATION OF INTEGRAL CALCULUS
Technological University of the Philippines
In this chapter of this course we will be looking at Applications of Integrals. There are many other
applications, however many of them require integration techniques. Because this chapter is focused
on the applications of integrals it is assumed in all the examples that you can do the integrals. There
1. be
will not Integrals
as muchasdetail
General andintegration
in the Particular Solution tothe
process in Differential
examplesEquation (Familyasofthere
in this chapter Curves)
was in the
2. Average Function Value
examples in the previous chapter.
3. Area Between Curves
Here is a listing of applications covered in this chapter.
4. Volume of Solid Revolution
5. Centroid
6. Arc length and Surface Area of a Curve
Integrals as General and Particular Solutions (Families of Curves)
The first-order equation 𝑑𝑦/𝑑𝑥 = 𝑓 (𝑥, 𝑦) takes an especially simple form if the right-hand-side
function f does not actually involve the dependent variable y, so 𝑑𝑦/𝑑𝑥 = 𝑓 (𝑥).
In this special case we need only integrate both sides to obtain 𝑦(𝑥) = 𝑓 (𝑥) 𝑑𝑥 + 𝐶. This is a general
solution, meaning that it involves an arbitrary constant 𝐶, and for every choice of C it is a solution of the
differential equation. If 𝐺(𝑥) is a particular antiderivative of 𝑓, that is, if 𝐺’(𝑥) ≡ 𝑓 (𝑥), then 𝑦(𝑥) =
𝐺(𝑥) + 𝐶. This equation is the General Solution.
The graphs of any two such solutions 𝑦1 (𝑥) = 𝐺(𝑥) + 𝐶1 and 𝑦2 (𝑥) = 𝐺(𝑥) + 𝐶2 interval I are
“parallel” in the sense as illustrated by the figure. There we see that the constant C is geometrically the
vertical distance between the two curves y(x) = G(x) and y(x) = G(x) + C. these parallel graphs are called
families of curves.
1
Figure. Graphs of 𝑦 = 𝑥 2 + 𝐶 for various values of C
4
To satisfy an initial condition 𝑦(𝑥0 ) = 𝑦0 , we need only substitute 𝑥 = 𝑥0 and 𝑦 = 𝑦0 into the
general solution to obtain 𝑦0 = 𝐺(𝑥0 ) + 𝐶, so that 𝐶 = 𝑦0 − 𝐺(𝑥0 ). With this choice of C, we
obtain the particular solution satisfying the initial value problem 𝑑𝑦/𝑑𝑥 = 𝑓 (𝑥), 𝑦(𝑥0 ) = 𝑦0 .
Example 1. Solve the initial value problem 𝑦′ = 2𝑥 + 3, 𝑦(1) = 2
𝑑𝑦
Solution. Recall that 𝑦 ′ = . And by using some algebra 𝑑𝑦 = (2𝑥 + 3)𝑑𝑥, integrate both side of the
𝑑𝑥
equation will yield to 𝑦 = 𝑥 2 + 3𝑥 + 𝐶. This is the general solution. To get particular solution by
,substitute 𝑥 = 1 and 𝑦 = 2 from the general solution 2 = (1)2 + 3(1) + 𝐶, thus 𝐶 = −2. Substituting C
from the general solution will give us 𝑦 = 𝑥 2 + 3𝑥 − 2.
Example 2. Solve the initial value problem 𝑦 ′ = 4𝑥 − 1, 𝑦(3) = 2
𝑑𝑦
Solution. Recall that 𝑦 ′ = . And by using some algebra 𝑑𝑦 = (4𝑥 − 1)𝑑𝑥, integrate both side of the
𝑑𝑥
equation will yield to 𝑦 = 2𝑥 2 − 𝑥 + 𝐶. This is the general solution. To get particular solution by
substitute 𝑥 = 3 and 𝑦 = 2 from the general solution 2 = 2(3)2 − (3) + 𝐶, thus 𝐶 = −13. Substituting
C from the general solution will give us 𝑦 = 2𝑥 2 − 𝑥 − 13.
Exercises. Find general solutions of the differential equation of the following and find the particular
solution using the given points.
1. 𝑦 ′ = 1 − 2𝑥, at point (−1, −1)
2. 𝑦 ′ = 𝑥 + 3𝑥 2 , at point (1, −1)
3. 𝑦 ′ = 4𝑥 − 3, at point (−2, 4)
4. 𝑥𝑦 ′ = 3𝑥 + 2, at point (2, 3)
5. 𝑥𝑦 ′ = 𝑥 − 5, at point (−3, 5)
6. 𝑦 ′ = 3𝑥𝑦 + 2𝑦, at point (1, 1)
7. x𝑦 ′ = 3𝑥𝑦 + 2𝑦, at point (2, 3)
Average Function Value
To find the average value of a set of numbers, you just add the numbers and divide by the number of
numbers. How would you find the average value of a continuous function over some interval?
The problem is that there are an infinite number of numbers to add up, then divide by infinity. One
approach is to divide up the interval and use n left or right samples of the value of the function, add
them up, then divide by n. If we take the limit as n approaches infinity, then we will get the average
value. The formula for the average value of a function, f, over the interval from a to b is:
𝑏
1
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = ∫ 𝑓(𝑥)𝑑𝑥
𝑏−𝑎 𝑎
Example. Determine the average value of each of the following functions on the given interval.
1. 𝑦 = 𝑥 2 + 1 at interval [−2, 2]
2. 𝑦 = 𝑥 3 at interval [−3, 3]
Solution:
Exercises
1. 𝑦 = 2𝑥 + 5 at interval [−10, 5]
2. 𝑦 = 𝑥 2 − 2𝑥 + 2 at interval [−2, 4]
𝑥+1
3. 𝑦 = at interval [1, 4]
𝑥
2𝑥−1
4. 𝑦 = 𝑒 at interval [−3, 2]
5. 𝑦 = sin 2𝑥 at interval [−𝜋, 𝜋]
Area Between Curves
In this section we are going to look at finding the area between two curves. There are two cases that we
are going to be looking at. In the first case we want to determine the area between 𝑦 = 𝑓(𝑥) and 𝑦 =
𝑔(𝑥) on the interval [𝑎, 𝑏]. We are also going to assume that 𝑓(𝑥) ≥ 𝑔(𝑥). Look at the following sketch
to get an idea of what we’re initially going to look at.
, The formula for the area in this case.
𝑏
𝐴 = ∫ [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
𝑎
The second case is almost identical to the first case. Here we are going to determine the area between
𝑥 = 𝑓(𝑦) and 𝑥 = 𝑔(𝑦) on the interval [𝑐, 𝑑] with 𝑓(𝑦) ≥ 𝑔(𝑦).
In this case the formula is,
𝑑
𝐴 = ∫ [𝑓(𝑦) − 𝑔(𝑦)]𝑑𝑦
𝑐
Now the two formulas are perfectly serviceable, however, it is sometimes easy to forget that these
always require the first function to be the larger of the two functions. So, instead of these formulas we
will instead use the following “word” formulas to make sure that we remember that the area is always
the “larger” function minus the “smaller” function.
In the first case we will use,
𝑏
𝐴 = ∫𝑎 [𝑢𝑝𝑝𝑒𝑟𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 − 𝑙𝑜𝑤𝑒𝑟𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛]𝑑𝑥, 𝑎 ≤ 𝑥 ≤ 𝑏
In the second case we will use,
𝑑
𝐴 = ∫𝑐 [𝑟𝑖𝑔ℎ𝑡𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 − 𝑙𝑒𝑓𝑡𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛]𝑑𝑦, 𝑐 ≤ 𝑦 ≤ 𝑑
APPLICATION OF INTEGRAL CALCULUS
Technological University of the Philippines
In this chapter of this course we will be looking at Applications of Integrals. There are many other
applications, however many of them require integration techniques. Because this chapter is focused
on the applications of integrals it is assumed in all the examples that you can do the integrals. There
1. be
will not Integrals
as muchasdetail
General andintegration
in the Particular Solution tothe
process in Differential
examplesEquation (Familyasofthere
in this chapter Curves)
was in the
2. Average Function Value
examples in the previous chapter.
3. Area Between Curves
Here is a listing of applications covered in this chapter.
4. Volume of Solid Revolution
5. Centroid
6. Arc length and Surface Area of a Curve
Integrals as General and Particular Solutions (Families of Curves)
The first-order equation 𝑑𝑦/𝑑𝑥 = 𝑓 (𝑥, 𝑦) takes an especially simple form if the right-hand-side
function f does not actually involve the dependent variable y, so 𝑑𝑦/𝑑𝑥 = 𝑓 (𝑥).
In this special case we need only integrate both sides to obtain 𝑦(𝑥) = 𝑓 (𝑥) 𝑑𝑥 + 𝐶. This is a general
solution, meaning that it involves an arbitrary constant 𝐶, and for every choice of C it is a solution of the
differential equation. If 𝐺(𝑥) is a particular antiderivative of 𝑓, that is, if 𝐺’(𝑥) ≡ 𝑓 (𝑥), then 𝑦(𝑥) =
𝐺(𝑥) + 𝐶. This equation is the General Solution.
The graphs of any two such solutions 𝑦1 (𝑥) = 𝐺(𝑥) + 𝐶1 and 𝑦2 (𝑥) = 𝐺(𝑥) + 𝐶2 interval I are
“parallel” in the sense as illustrated by the figure. There we see that the constant C is geometrically the
vertical distance between the two curves y(x) = G(x) and y(x) = G(x) + C. these parallel graphs are called
families of curves.
1
Figure. Graphs of 𝑦 = 𝑥 2 + 𝐶 for various values of C
4
To satisfy an initial condition 𝑦(𝑥0 ) = 𝑦0 , we need only substitute 𝑥 = 𝑥0 and 𝑦 = 𝑦0 into the
general solution to obtain 𝑦0 = 𝐺(𝑥0 ) + 𝐶, so that 𝐶 = 𝑦0 − 𝐺(𝑥0 ). With this choice of C, we
obtain the particular solution satisfying the initial value problem 𝑑𝑦/𝑑𝑥 = 𝑓 (𝑥), 𝑦(𝑥0 ) = 𝑦0 .
Example 1. Solve the initial value problem 𝑦′ = 2𝑥 + 3, 𝑦(1) = 2
𝑑𝑦
Solution. Recall that 𝑦 ′ = . And by using some algebra 𝑑𝑦 = (2𝑥 + 3)𝑑𝑥, integrate both side of the
𝑑𝑥
equation will yield to 𝑦 = 𝑥 2 + 3𝑥 + 𝐶. This is the general solution. To get particular solution by
,substitute 𝑥 = 1 and 𝑦 = 2 from the general solution 2 = (1)2 + 3(1) + 𝐶, thus 𝐶 = −2. Substituting C
from the general solution will give us 𝑦 = 𝑥 2 + 3𝑥 − 2.
Example 2. Solve the initial value problem 𝑦 ′ = 4𝑥 − 1, 𝑦(3) = 2
𝑑𝑦
Solution. Recall that 𝑦 ′ = . And by using some algebra 𝑑𝑦 = (4𝑥 − 1)𝑑𝑥, integrate both side of the
𝑑𝑥
equation will yield to 𝑦 = 2𝑥 2 − 𝑥 + 𝐶. This is the general solution. To get particular solution by
substitute 𝑥 = 3 and 𝑦 = 2 from the general solution 2 = 2(3)2 − (3) + 𝐶, thus 𝐶 = −13. Substituting
C from the general solution will give us 𝑦 = 2𝑥 2 − 𝑥 − 13.
Exercises. Find general solutions of the differential equation of the following and find the particular
solution using the given points.
1. 𝑦 ′ = 1 − 2𝑥, at point (−1, −1)
2. 𝑦 ′ = 𝑥 + 3𝑥 2 , at point (1, −1)
3. 𝑦 ′ = 4𝑥 − 3, at point (−2, 4)
4. 𝑥𝑦 ′ = 3𝑥 + 2, at point (2, 3)
5. 𝑥𝑦 ′ = 𝑥 − 5, at point (−3, 5)
6. 𝑦 ′ = 3𝑥𝑦 + 2𝑦, at point (1, 1)
7. x𝑦 ′ = 3𝑥𝑦 + 2𝑦, at point (2, 3)
Average Function Value
To find the average value of a set of numbers, you just add the numbers and divide by the number of
numbers. How would you find the average value of a continuous function over some interval?
The problem is that there are an infinite number of numbers to add up, then divide by infinity. One
approach is to divide up the interval and use n left or right samples of the value of the function, add
them up, then divide by n. If we take the limit as n approaches infinity, then we will get the average
value. The formula for the average value of a function, f, over the interval from a to b is:
𝑏
1
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = ∫ 𝑓(𝑥)𝑑𝑥
𝑏−𝑎 𝑎
Example. Determine the average value of each of the following functions on the given interval.
1. 𝑦 = 𝑥 2 + 1 at interval [−2, 2]
2. 𝑦 = 𝑥 3 at interval [−3, 3]
Solution:
Exercises
1. 𝑦 = 2𝑥 + 5 at interval [−10, 5]
2. 𝑦 = 𝑥 2 − 2𝑥 + 2 at interval [−2, 4]
𝑥+1
3. 𝑦 = at interval [1, 4]
𝑥
2𝑥−1
4. 𝑦 = 𝑒 at interval [−3, 2]
5. 𝑦 = sin 2𝑥 at interval [−𝜋, 𝜋]
Area Between Curves
In this section we are going to look at finding the area between two curves. There are two cases that we
are going to be looking at. In the first case we want to determine the area between 𝑦 = 𝑓(𝑥) and 𝑦 =
𝑔(𝑥) on the interval [𝑎, 𝑏]. We are also going to assume that 𝑓(𝑥) ≥ 𝑔(𝑥). Look at the following sketch
to get an idea of what we’re initially going to look at.
, The formula for the area in this case.
𝑏
𝐴 = ∫ [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
𝑎
The second case is almost identical to the first case. Here we are going to determine the area between
𝑥 = 𝑓(𝑦) and 𝑥 = 𝑔(𝑦) on the interval [𝑐, 𝑑] with 𝑓(𝑦) ≥ 𝑔(𝑦).
In this case the formula is,
𝑑
𝐴 = ∫ [𝑓(𝑦) − 𝑔(𝑦)]𝑑𝑦
𝑐
Now the two formulas are perfectly serviceable, however, it is sometimes easy to forget that these
always require the first function to be the larger of the two functions. So, instead of these formulas we
will instead use the following “word” formulas to make sure that we remember that the area is always
the “larger” function minus the “smaller” function.
In the first case we will use,
𝑏
𝐴 = ∫𝑎 [𝑢𝑝𝑝𝑒𝑟𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 − 𝑙𝑜𝑤𝑒𝑟𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛]𝑑𝑥, 𝑎 ≤ 𝑥 ≤ 𝑏
In the second case we will use,
𝑑
𝐴 = ∫𝑐 [𝑟𝑖𝑔ℎ𝑡𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 − 𝑙𝑒𝑓𝑡𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛]𝑑𝑦, 𝑐 ≤ 𝑦 ≤ 𝑑
