5.4. Applications of integrals
Example 5.39 A company’s marginal cost function is given by MC(q) = 2q + 100 eq,
and its fixed costs are 10, 000. What is the cost function, C(q), for this company?
Using (5.2) above, we see that the cost function is given by the integral of the marginal cost, i.e.
C(q) = Z
(2q + 100 eq) dq = q2+ 100 eq+c,
where c is an arbitrary constant. This tells us, depending on the value of c, all of the possible cost functions for this company. But, which one should we take? Obviously, perhaps, we want the one which also gives us fixed costs of 10, 000, i.e. we want C(0) = 10, 000 =⇒ 10, 000 = 02+ 100 e0+c =⇒ 10, 000 = 100 + c =⇒ c = 9, 900, as the fixed costs are the cost of producing nothing. Thus, the cost function for this company is given by
C(q) = q2+ 100 eq+9, 900,
as this function agrees with the question on both the marginal and the fixed costs of production.
5.4.2 Consumer and producer surpluses
Suppose that a market has linear supply and demand functions as illustrated in Figure 5.8. As we know from Section 2.1.5, the equilibrium price, p∗, and the equilibrium quantity, q∗, occur at the point where the graphs of these functions
intersect. Indeed, at equilibrium, as the consumers buy q∗ units of the good at a price of p∗ per unit, they pay an amount p∗q∗ to the suppliers and we can think of this as the area of the hatched region in Figure 5.9(b).
However, if the consumers are willing to buy q∗ units of the good, it can be argued7 that the consumers would be willing to pay an amount given by
Z q∗ 0
pD(q) dq,
which is the area of the hatched region in Figure 5.9(a). The difference between the area that represents what they would pay and the area that represents what they actually pay, i.e. the area of the hatched region in Figure 5.9(d), is called the consumer surplus.
Indeed, this consumer surplus, CS, can be found using the formula CS =
Z q∗ 0
pD(q) dq− p∗q∗,
and this is the amount that the consumers save by paying what they actually paid instead of what they would have paid.
7See, for example, Section 25.1 of Anthony and Biggs (1996).
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p
q S
D
O q∗
p∗
Figure 5.8:Linear supply and demand functions for a market. Note that the equilibrium price, p∗, and the equilibrium quantity, q∗, occur at the point where the graphs of these functions intersect.
Similarly, if the suppliers are willing to supply q∗ units of the good, it can be argued that they need to be paid an amount given by
Z q∗ 0
pS(q) dq,
which is the area of the hatched region in Figure 5.9(c). The difference between the area that represents what they are actually paid and the area that represents what they need to be paid, i.e. the area of the hatched region in Figure 5.9(e), is called the producer surplus. Indeed, this producer surplus, PS, can be found using the formula
PS = p∗q∗ − Z q∗
0
pD(q) dq,
and this is the amount that the suppliers gain by being paid what they actually receive instead of what they need to receive. Let’s look at a simple example.
Example 5.40 A market has an inverse demand function given by pD(q) = 70− 1
3q, and an inverse supply function given by
pS(q) = 20 +1 2q.
Find the equilibrium price and quantity. What are the consumer and producer surpluses for this market?
The equilibrium quantity, q∗, makes the prices obtained from the inverse demand and supply functions equal, i.e.
70− 1
3q∗ = 20 + 1
2q∗ =⇒ 50 = 5
6q∗ =⇒ q∗ = 60,
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Figure 5.9:What people pay or need to be paid. (a) What the consumers would pay for a quantity q∗. (b) What the consumers pay for a quantity q∗ if the market is at equilibrium.
(c) What the suppliers need to be paid for a quantity q∗. (d) What the consumers save if they pay for a quantity q∗ in a market that is at equilibrium, this is the consumer surplus.
(e) What the producers gain if they sell a quantity q∗ in a market that is at equilibrium, this is the producer surplus.
and this means that the equilibrium price, p∗, is given by p∗ = 70− 1
3(60) = 70− 20 = 50, if we use the inverse demand function.
Hence, to find the consumer surplus, CS, we have CS =
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is the consumer surplus. And, to find the producer surplus, PS, we have PS = p∗q∗−
Z q∗ 0
pS(q) dq, and so we need to find
Z 60 0
20 + 1 2q
dq =
20q +q2 4
60 0
=
20(60) + 1 4(60)2
− 0 = 1, 200 + 900 = 2, 100, which means that
PS = (50)(60)− 2, 100 = 3, 000 − 2, 100 = 900, is the producer surplus.
Although, as both the demand and supply functions are linear in this example, there is an easier way to find the consumer and producer surpluses as the next Activity shows.
Activity 5.21 Sketch the inverse demand and supply functions in the previous example and shade in the regions which represent the consumer and producer surplus. What are the areas of these regions?
Of course, the demand and supply functions that we are given may not be linear and, in such cases, we would have to use integration to find the consumer and producer
surpluses.
Activity 5.22 The demand for a commodity is given by the equation p(q + 1) = 231.
If the equilibrium quantity is 10, find the equilibrium price and hence determine the consumer surplus.
Learning outcomes
At the end of this chapter and having completed the relevant reading and activities, you should be able to:
find integrals using standard integrals and the rules of integration;
find integrals by simplifying the integrand using partial fractions and trigonometric identities;
use integrals to find areas;
solve problems from economics-based subjects that involve integrals.
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5.4. Solutions to activities
Solutions to activities
Solution to activity 5.1
Given the linear combination rule, i.e.
Z we can derive the constant multiple rule by setting l = 0 so that
Z
Suppose that F (x) and G(x) are antiderivatives of f (x) and g(x) respectively, i.e.
dF
where c is an arbitrary constant. But, by the linear combination rule for differentiation, we also have as they have the same antiderivatives.
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Solution to activity 5.3
For (a), use the constant multiple rule to see that Z
−3 cos x dx = −3 Z
cos x dx =−3 sin x + c, where c is an arbitrary constant. For (b), we use the sum rule to see that
Z
where c is an arbitrary constant. For (c), we use the linear combination rule to see that Z where c is an arbitrary constant.
Solution to activity 5.4
For both of these integrals we use the substitution g = 4x + 7 so that we have dg
dx = 4 =⇒ dg = 4dx =⇒ dx = 1 4dg.
Hence making this substitution in the first integral we get
Z 1 where c is an arbitrary constant whereas, in the second integral, we get
Z where c is an arbitrary constant.
Solution to activity 5.5
Using the standard integrals as a source of antiderivatives, we see that, if n6= −1, Z where c is an arbitrary constant. We also have
Z
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5.4. Solutions to activities
Z
cos(ax + b) dx = 1
asin(ax + b) + c, where c is an arbitrary constant.
Of course, if a = 0, then the dependence on x in the integrand disappears and so we are just integrating a constant, i.e. we have
Z where c is an arbitrary constant.
Solution to activity 5.6
Using what we saw in Activity 5.5 we see that the integrals from Activity 5.4 are, simply, where c is an arbitrary constant. This is, of course, exactly what we found in Activity 5.4.
Solution to activity 5.7
Taking g = 3x2+ 7 we have g0(x) = 6x and so dg = 6x dx, i.e. x dx = 16 dg. Hence, in the first integral, this substitution gives
Z x where c is an arbitrary constant whereas, in the second integral, this substitution gives
Z
where c is an arbitrary constant. In both cases, note that the extra ‘x’ in the integrand was actually needed for the substitution g = 3x2+ 7 to work.
Solution to activity 5.8
Here the composition is sin(x2) and so we take g = x2. As such, we have dg
dx = 2x =⇒ x dx = 1
2dg,
which is a constant multiple of the other part of the product in the integrand, i.e. this substitution will work. Thus, the substitution gives
Z where c is an arbitrary constant. Here, of course, the extra ‘x’ in the integrand was needed for the substitution g = x2 to work.
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Solution to activity 5.9
Here the composition is cos2x and so we take g = cos x. As such, we have dg
dx =− sin x,
which, up to a minus, is the other part of the product in the integrand, i.e. this substitution will work. Thus, we see that
dg =− sin x dx, and so the substitution gives
Z
sin2x cos x dx = Z
g2(− dg) = − Z
g2dg = −g3
3 + c =−1
3cos3x + c.
Here, of course, the extra ‘sin x’ in the integrand was needed for the substitution g = cos x to work.
Solution to activity 5.10 In Activity 2.4, we saw that
cot x = cos x sin x,
which means that the composition is (sin x)−1 and so we take g = sin x. As such, we
have dg
dx = cos x,
which is the other part of the product in the integrand, i.e. this substitution will work.
Thus, we see that
dg = cos x dx, and so the substitution gives
Z
cot x dx =
Z cos x sin xdx =
Z dg
g = ln|g| + c = ln | sin x| + c.
Here, of course, the extra ‘cos x’ in the integrand was needed for the substitution g = sin x to work.
Solution to activity 5.11
We note that the quadratic expression in the denominator can be written as x2 + 2x + 2 = (x + 1)2+ 1,
if we complete the square. As such, we have
Z dx
x2+ 2x + 2 =
Z dx
(x + 1)2+ 1 = tan−1(x + 1) + c,
using the result we derived in Example 5.17. (A useful exercise at this point is to try and get this answer by actually making the substitution x + 1 = tan θ as we did in that example.)
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5.4. Solutions to activities
Solution to activity 5.12
Using the change of base formula for logarithms from Section 2.1.4, i.e.
loga(x) = ln x where c is an arbitrary constant.
Solution to activity 5.13 To find
Z
x ln x dx the other way, i.e. by choosing
f (x) = x and g0(x) = ln x,
we differentiate f (x) and integrate g0(x) using the result in Example 5.20 to get f0(x) = 1 and g0(x) = x ln x− x,
where we have suppressed the arbitrary constant from the integration. Applying the rule then gives
so that, taking the integral on the right-hand-side over to the left-hand-side, we have 2 where c is an arbitrary constant. Notice that this is the same as the answer we found in Example 5.19 but it is slightly trickier to get and we need to know the answer to
Example 5.20.
Solution to activity 5.14
Unlike what we saw in Example 5.21, it would actually make more sense to find Z
(x2+ 1)2x2dx,
by multiplying out the brackets and integrating term-by-term rather than integrating it by parts. Doing this, we get
Z
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where c is an arbitrary constant. Indeed, to verify that this is the same answer as the one we saw in the example, it is easiest to take the earlier answer and note that
x3 which is what we got above.
Solution to activity 5.15
To find this integral we also use the other double-angle formula from Activity 2.18, namely
as this allows us to write the problematic integrand cos2x in terms of the function cos(2x) which is far easier to integrate. This means that we have
Z where c is an arbitrary constant.
Solution to activity 5.16
Using the first step, we can see that Z b
as F (x) + c is also an antiderivative of f (x) if c is a constant. Then, using the second step we get8
which is exactly what we wanted. That is, including a constant of integration does not affect the value of a definite integral and so we can omit it.
Solution to activity 5.17
For definite integrals, it should be easy to see that we have the
constant multiple rule: If k is a constant and f (x) is a function, then Z b