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Applied Mathematics for Class 11th & 12th (Concepts and Questions)
11th Concepts Questions
12th Concepts Questions

Applied Maths Class 12th Chapters (Concepts)
1. Numbers, Quantification and Numerical Applications 2. Matrices 3. Differentiation and Its Applications
4. Integration and Its Application 5. Differential Equations and Modeling 6. Probability Distribution
7. Inferential Statistics 8. Index Numbers and Time Based Data 9. Financial Mathematics
10. Linear Programming

Content On This Page
Integration Indefinite Integrals as Family of Curves Definite Integrals as Area under the Curve
Application of Integration


Chapter 4 Integration and Its Application (Concepts)

Welcome to this essential chapter introducing the fundamental concepts of Integral Calculus within the framework of Applied Mathematics. While differential calculus focuses on rates of change and slopes, integral calculus provides the tools for the reverse process – anti-differentiation – and, crucially, for calculating the accumulation of quantities, such as areas under curves, total change from a rate, and economic surpluses. Understanding integration is vital for modeling and solving problems where we know the rate at which something changes and need to determine the total amount, a common requirement in economics, finance, physics, and engineering. This chapter will guide you through the core ideas of indefinite and definite integrals, basic integration techniques, and significant applications, particularly in economic analysis.

We begin with the concept of the Indefinite Integral, which seeks to find the family of functions whose derivative is a given function $f(x)$. This is denoted by $\int f(x) dx = F(x) + C$, where $F(x)$ is an anti-derivative (meaning $F'(x) = f(x)$) and $C$ is the arbitrary constant of integration, representing the family of all possible anti-derivatives. We will focus on mastering the basic integration formulas derived directly from standard differentiation rules:

While advanced techniques exist, we will explore foundational methods like integration by substitution, where we identify a suitable part of the integrand $u=g(x)$ such that its derivative $g'(x)$ is also present (or can be easily introduced), allowing us to transform the integral into a simpler form in terms of $u$. This significantly extends the applicability of our basic formulas.

Moving from the family of anti-derivatives, we introduce the Definite Integral, denoted by $\int\limits_{a}^{b} f(x) dx$. This represents the net accumulation of the quantity represented by $f(x)$ as $x$ ranges from a lower limit $a$ to an upper limit $b$. Geometrically, it is often interpreted as the signed area under the curve $y=f(x)$ between $x=a$ and $x=b$. The primary tool for evaluating definite integrals is the Fundamental Theorem of Calculus (Part II), which elegantly connects differentiation and integration: if $F'(x) = f(x)$, then $\int\limits_{a}^{b} f(x) dx = F(b) - F(a)$. This theorem allows us to calculate definite integrals by finding an anti-derivative and evaluating it at the limits of integration.

The true power of integration in Applied Mathematics shines through its applications. A key focus is on economic applications:

This chapter equips you with the foundational techniques of integral calculus and demonstrates their utility in translating rates of change back into total amounts, providing crucial tools for quantitative analysis in economics, finance, and beyond.



Integration


Integration is a fundamental concept in calculus and can be viewed as the inverse process of differentiation. While differentiation helps us find the instantaneous rate of change or the slope of a tangent line, integration helps us find the original function given its rate of change, or compute the area under a curve.

If we are given a function $f(x)$ and we need to find a function $F(x)$ such that the derivative of $F(x)$ is equal to $f(x)$, i.e., $\frac{d}{dx} F(x) = f(x)$, then $F(x)$ is called an antiderivative or a primitive of $f(x)$. The process of finding such a function is known as integration.

The symbol $\int$ is used to denote integration. It is an elongated 'S', originally representing a sum, connecting integration to the concept of summing up infinitesimally small quantities (like areas of rectangles under a curve, which we'll see in definite integrals). The expression $\int f(x) \, dx$ represents the indefinite integral of $f(x)$ with respect to $x$.

If $\frac{d}{dx} F(x) = f(x)$, then $\int f(x) \, dx = F(x) + C$

... (i)

Here, $F(x)$ is any particular antiderivative of $f(x)$, and $C$ is the constant of integration. The term 'indefinite' indicates that the result is not a single function but a family of functions, all differing by a constant.

Why the Constant of Integration?

Consider finding the antiderivative of $f(x) = 2x$. We know that $\frac{d}{dx}(x^2) = 2x$. So, $F(x) = x^2$ is an antiderivative. However, let's also consider the derivatives of $x^2 + 5$, $x^2 - 7$, or $x^2 + \pi$:

$\frac{d}{dx}(x^2) = 2x$

$\frac{d}{dx}(x^2 + 5) = 2x$

$\frac{d}{dx}(x^2 - 7) = 2x$

$\frac{d}{dx}(x^2 + \pi) = 2x$

In all these cases, the derivative is $2x$. This means that when we integrate $2x$, the antiderivative could be $x^2$, $x^2+5$, $x^2-7$, $x^2+\pi$, or $x^2$ plus any other constant. Thus, the indefinite integral of $2x$ is written as $\int 2x \, dx = x^2 + C$, where $C$ can be any real number. This constant $C$ accounts for all possible constant terms that would vanish upon differentiation.


Properties of Indefinite Integrals

Indefinite integrals follow certain properties that make calculations easier, similar to the properties of differentiation.

1. Linearity Property: Integration is a linear operation.

These two parts can be combined into a single property:

$\int [k_1 f(x) \pm k_2 g(x)] \, dx = k_1 \int f(x) \, dx \pm k_2 \int g(x) \, dx$

2. Relationship with Differentiation:


Basic Integration Formulas

Here is a list of fundamental indefinite integration formulas, which are the inverse of the basic differentiation rules. It is essential to remember these formulas.

Function $f(x)$ Indefinite Integral $\int f(x) \, dx$
$x^n$ $\frac{x^{n+1}}{n+1} + C$, for $n \ne -1$
$\frac{1}{x}$ $\log |x| + C$, for $x \ne 0$
$e^x$ $e^x + C$
$a^x$ $\frac{a^x}{\log a} + C$, for $a > 0, a \ne 1$
$\sin x$ $-\cos x + C$
$\cos x$ $\sin x + C$
$\tan x$ $\log |\sec x| + C$ or $-\log |\cos x| + C$
$\cot x$ $\log |\sin x| + C$
$\sec x$ $\log |\sec x + \tan x| + C$ or $\log \left|\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right| + C$
$\text{cosec} x$ $\log |\text{cosec} x - \cot x| + C$ or $\log \left|\tan\left(\frac{x}{2}\right)\right| + C$
$\sec^2 x$ $\tan x + C$
$\text{cosec}^2 x$ $-\cot x + C$
$\sec x \tan x$ $\sec x + C$
$\text{cosec} x \cot x$ $-\text{cosec} x + C$
Constant $k$ $kx + C$

Note the inclusion of the absolute value in the arguments of the logarithm function. For example, $\int \frac{1}{x} \, dx = \log |x| + C$. This is because the function $f(x) = \frac{1}{x}$ is defined for all $x \ne 0$, while the function $\log x$ is only defined for $x > 0$. The antiderivative must be defined on the same domain as the original function. $\log|x|$ handles both positive and negative $x$ (except $x=0$).


Methods of Integration

While the basic formulas allow us to integrate simple functions, many functions require specific techniques to find their antiderivatives. The main methods of integration include:

1. Integration by Substitution (or Change of Variable)

This is one of the most common and powerful techniques. It is based on the reverse of the chain rule for differentiation. The goal is to transform the integral into a simpler form by introducing a new variable, say $u$, where $u$ is a function of the original variable. The key is to identify a part of the integrand that can be chosen as $u$ such that the differential $du$ (which is $du = \frac{du}{dx} \, dx$) is also present in the integrand (perhaps multiplied by a constant).
The idea is: if $\int f(u) \, du = F(u) + C$, then $\int f(g(x)) g'(x) \, dx = F(g(x)) + C$. By substituting $u = g(x)$ and $du = g'(x) \, dx$, the complex integral on the right becomes the simpler integral on the left.

2. Integration by Parts

This technique is used to integrate the product of two functions. It is derived from the product rule for differentiation. The formula is:

$\mathbf{\int u \, dv = uv - \int v \, du}$

... (vi)

Here, $u$ and $v$ are functions of the variable of integration. We need to split the integrand into two parts: one part is chosen as $u$ (which will be differentiated to find $du$), and the remaining part is chosen as $dv$ (which will be integrated to find $v$). The goal is to make the new integral $\int v \, du$ simpler to evaluate than the original integral $\int u \, dv$. A common mnemonic to help choose which part to set as $u$ is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), prioritizing functions earlier in this list as $u$.

3. Integration by Partial Fractions

This method is used specifically for integrating rational functions, i.e., functions of the form $\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials. If the degree of the numerator is less than the degree of the denominator, we decompose the rational function into a sum of simpler fractions called partial fractions. The form of these partial fractions depends on the factorization of the denominator $Q(x)$ into linear and irreducible quadratic factors. Each of these simpler fractions can then be integrated using basic formulas or substitution. If the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$, we first perform polynomial long division to write $\frac{P(x)}{Q(x)}$ as a polynomial plus a proper rational function.


Examples

Example 1. Find the indefinite integral of $f(x) = 3x^2 + 4x - 5$.

Answer:

We need to evaluate the indefinite integral $\int (3x^2 + 4x - 5) \, dx$.

Using the linearity property [Equations (ii) and (iii)]:

$\int (3x^2 + 4x - 5) \, dx = \int 3x^2 \, dx + \int 4x \, dx - \int 5 \, dx$

$= 3 \int x^2 \, dx + 4 \int x \, dx - 5 \int 1 \, dx$

Now, apply the power rule $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ and the constant rule $\int k \, dx = kx + C$:

$= 3 \left(\frac{x^{2+1}}{2+1}\right) + 4 \left(\frac{x^{1+1}}{1+1}\right) - 5 (x) + C$

$= 3 \left(\frac{x^3}{3}\right) + 4 \left(\frac{x^2}{2}\right) - 5x + C$

Simplify the terms:

$= \cancel{3} \frac{x^3}{\cancel{3}} + \cancel{4}^{2} \frac{x^2}{\cancel{2}_{1}} - 5x + C$

[Cancellation]

$= x^3 + 2x^2 - 5x + C$

... (i)

The indefinite integral of $3x^2 + 4x - 5$ is $x^3 + 2x^2 - 5x + C$.

To verify, we can differentiate the result: $\frac{d}{dx}(x^3 + 2x^2 - 5x + C) = 3x^2 + 4x - 5 + 0 = 3x^2 + 4x - 5$, which is the original function.



Indefinite Integrals as Family of Curves


As we discussed in the previous section, the indefinite integral of a function $f(x)$, denoted by $\int f(x) \, dx$, represents the collection of all possible antiderivatives of $f(x)$. If $F(x)$ is one particular antiderivative (meaning $\frac{d}{dx} F(x) = f(x)$), then any function of the form $F(x) + C$, where $C$ is an arbitrary constant, is also an antiderivative.

$\int f(x) \, dx = F(x) + C$

This arbitrary constant $C$, known as the constant of integration, has a significant geometric interpretation. When we consider the graph of the indefinite integral $y = F(x) + C$, each specific value assigned to $C$ corresponds to a particular curve in the coordinate plane. As $C$ varies over all real numbers, the equation $y = F(x) + C$ generates an entire family of curves.

Geometrically, all the curves in this family $y = F(x) + C$ are vertical translations of each other. The graph of $y = F(x) + C$ is simply the graph of $y = F(x)$ shifted upwards by $C$ units if $C > 0$, or downwards by $|C|$ units if $C < 0$. All these curves share the same shape but occupy different positions vertically.

Illustrative Example: The Family of Curves for $\int 2x \, dx$

Consider the indefinite integral of the function $f(x) = 2x$.

$\int 2x \, dx = x^2 + C$

Here, $F(x) = x^2$ is a particular antiderivative. The indefinite integral $y = x^2 + C$ represents a family of parabolas.

The graphs of these functions are parabolas opening upwards, all having the same shape (determined by the $x^2$ term) but differing in their vertical position (determined by the constant $C$).

Graph showing parabolas y=x^2, y=x^2+1, y=x^2-2, etc., representing a family of curves

The geometric significance of $\frac{d}{dx} (F(x) + C) = f(x)$ for all values of $C$ means that for any given value of $x$, the slope of the tangent line to every curve in the family $y = F(x) + C$ is the same, and this slope is equal to $f(x)$. If you draw a vertical line $x = x_0$, all the points where this vertical line intersects the curves $y = F(x) + C$ will have tangent lines that are parallel to each other, with a common slope of $f(x_0)$.

Finding a Particular Curve from the Family (Initial Value Problem)

While the indefinite integral gives a family of curves, in many applications (like physics or economics), we are interested in a specific curve that satisfies certain conditions. If we are given the rate of change of a quantity ($f(x)$) and the value of the quantity at a specific point (an initial condition), we can use this information to determine the unique value of the constant $C$ and thus identify the particular curve from the family that satisfies the given condition.

If we are given that the curve $y = F(x) + C$ passes through a specific point $(x_0, y_0)$, we can substitute these values into the equation to find $C$:

$y_0 = F(x_0) + C$

Solving for $C$:

$C = y_0 - F(x_0)$

Substituting this value of $C$ back into $y = F(x) + C$ gives the equation of the unique curve that satisfies the given initial condition. Problems of this type, where the derivative is given along with a point on the original function, are called initial value problems.


Example

Example 1. Find the equation of the curve whose slope at any point $(x, y)$ is given by $\frac{dy}{dx} = 3x^2 - 4$, and which passes through the point $(1, 5)$.

Answer:

Given that the slope of the curve at any point $(x, y)$ is $\frac{dy}{dx} = 3x^2 - 4$. This is the derivative of the function representing the curve.

$\frac{dy}{dx} = 3x^2 - 4$

To find the equation of the curve $y = f(x)$, we need to integrate the given derivative:

$y = \int (3x^2 - 4) \, dx$

Using the linearity property and the power rule of integration:

$y = \int 3x^2 \, dx - \int 4 \, dx$

$y = 3 \int x^2 \, dx - 4 \int 1 \, dx$

$y = 3 \left(\frac{x^{2+1}}{2+1}\right) - 4(x) + C$

$y = 3 \left(\frac{x^3}{3}\right) - 4x + C$

$y = x^3 - 4x + C$

... (i)

This equation $y = x^3 - 4x + C$ represents the family of curves whose slope is $3x^2 - 4$.

We are given the additional condition that the curve passes through the point $(1, 5)$. This means that when $x=1$, $y=5$. We substitute these values into equation (i) to find the specific value of $C$.

$5 = (1)^3 - 4(1) + C$

[Substituting $x=1, y=5$ into (i)]

$5 = 1 - 4 + C$

$5 = -3 + C$

Solving for $C$:

$C = 5 + 3 = 8$

... (ii)

Substitute the value of $C=8$ back into the equation of the family of curves (i):

$\mathbf{y = x^3 - 4x + 8}$

This is the equation of the specific curve that satisfies both the given slope condition and the initial condition of passing through the point $(1, 5)$.



Definite Integrals as Area under the Curve


In contrast to indefinite integrals, which represent a family of functions, definite integrals evaluate to a specific numerical value. This value has profound interpretations, one of the most intuitive being the area of a region in the plane. Specifically, the definite integral of a non-negative function over an interval represents the area of the region bounded by the graph of the function, the x-axis, and the vertical lines at the endpoints of the interval.


Definition of Definite Integral (Conceptual Link to Area)

Imagine a function $f(x)$ that is continuous and non-negative ($f(x) \ge 0$) on a closed interval $[a, b]$. We want to find the area of the region situated between the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$.

The fundamental idea behind the definite integral is to approximate this area by dividing the interval $[a, b]$ into a large number of small subintervals. On each subinterval, we construct a rectangle whose height is determined by the function's value at some point within that subinterval (e.g., the left endpoint, right endpoint, or midpoint) and whose width is the length of the subinterval. The sum of the areas of these rectangles provides an approximation of the area under the curve. This sum is known as a Riemann sum.

Graph showing a curve with rectangles approximating the area under it

As we increase the number of subintervals ($n \to \infty$) and, consequently, the width of each subinterval approaches zero ($\Delta x \to 0$), the approximation becomes more accurate. The definite integral is formally defined as the limit of these Riemann sums:

$\int\limits_a^b f(x) \, dx = \lim\limits_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$

... (i)

Here, $a$ and $b$ are the lower and upper limits of integration, respectively. The interval $[a, b]$ is partitioned into $n$ subintervals, $\Delta x$ is the width of each subinterval (if they are equal width), and $x_i^*$ is any point chosen within the $i$-th subinterval.


Fundamental Theorem of Calculus (Part 2)

The definition involving Riemann sums provides the theoretical basis for the definite integral as area. However, evaluating definite integrals directly using this limit definition is cumbersome. The Fundamental Theorem of Calculus (Part 2) provides a much more efficient way to evaluate definite integrals by connecting them to antiderivatives.

It states that if $f(x)$ is a continuous function on the closed interval $[a, b]$, and if $F(x)$ is any antiderivative of $f(x)$ on that interval (i.e., $F'(x) = f(x)$ for all $x$ in $[a, b]$), then the definite integral of $f(x)$ from $a$ to $b$ can be evaluated by finding the difference between the value of the antiderivative at the upper limit ($b$) and the value of the antiderivative at the lower limit ($a$).

$\mathbf{\int\limits_a^b f(x) \, dx = [F(x)]_a^b = F(b) - F(a)}$

... (ii)

When using this theorem, we do not need to include the constant of integration $C$ from the indefinite integral because it cancels out: $[F(x) + C]_a^b = (F(b) + C) - (F(a) + C) = F(b) + C - F(a) - C = F(b) - F(a)$.


Interpretation of Definite Integral as Area

The interpretation of the definite integral $\int_a^b f(x) \, dx$ as area depends on the sign of the function $f(x)$ over the interval $[a, b]$.


Example of Definite Integral Calculation

Example 1. Evaluate the definite integral $\int\limits_1^2 x^2 \, dx$. Interpret the result in terms of area.

Answer:

We need to evaluate $\int\limits_1^2 x^2 \, dx$. The function is $f(x) = x^2$, and the interval of integration is $[1, 2]$.

Step 1: Find an antiderivative $F(x)$ of $f(x) = x^2$. Using the power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$:

$\int x^2 \, dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C$

For the Fundamental Theorem of Calculus, we can choose any antiderivative, so we typically take the one with $C=0$. Let $F(x) = \frac{x^3}{3}$.

Step 2: Use the Fundamental Theorem of Calculus (Part 2) to evaluate the definite integral. The lower limit is $a=1$ and the upper limit is $b=2$.

$\int\limits_1^2 x^2 \, dx = [F(x)]_1^2 = \left[\frac{x^3}{3}\right]_1^2$

[Applying Theorem (ii)]

Now, evaluate $F(b) - F(a)$:

$= F(2) - F(1)$

$= \frac{(2)^3}{3} - \frac{(1)^3}{3}$

$= \frac{8}{3} - \frac{1}{3}$

$= \frac{8-1}{3} = \frac{7}{3}$

The value of the definite integral is $\frac{7}{3}$.

Interpretation in terms of area: The function $f(x) = x^2$ is a parabola opening upwards. On the interval $[1, 2]$, for any $x$ between 1 and 2 (inclusive), $x^2$ is positive ($\ge 0$). Therefore, the value of the definite integral $\int\limits_1^2 x^2 \, dx = \frac{7}{3}$ represents the area of the region bounded by the curve $y = x^2$, the x-axis, and the vertical lines $x = 1$ and $x = 2$.

The area under the curve $y=x^2$ from $x=1$ to $x=2$ is $\frac{7}{3}$ square units.



Application of Integration


Integration is a fundamental operation in calculus with a wide range of applications across various disciplines, including physics, engineering, economics, statistics, and probability. Its utility stems primarily from two key interpretations: as the process of finding a total quantity from its rate of change (antidifferentiation) and as a method for calculating areas under curves. In Applied Mathematics, particularly in business and economics, integration is extensively used for concepts like finding total cost, revenue, and profit from their marginal functions, and for calculating consumer and producer surplus.


Finding Total Quantity from Marginal Function

We have previously established that marginal cost, marginal revenue, and marginal profit functions are the first derivatives of the total cost, total revenue, and total profit functions, respectively. Conversely, this means that the total function can be obtained by integrating its corresponding marginal function. This is a direct application of the concept of indefinite integrals as antiderivatives.

Total Cost Function from Marginal Cost Function

If $MC(x)$ represents the marginal cost function (the rate of change of total cost with respect to the number of units produced, $x$), then the total cost function, $C(x)$, is the indefinite integral of $MC(x)$ with respect to $x$.

$C(x) = \int MC(x) \, dx + K$

... (i)

Here, $K$ is the constant of integration. In the context of cost functions, this constant $K$ represents the fixed cost – the cost incurred by the firm even when zero units are produced. That is, $C(0) = K$. The value of $K$ can be determined if either the fixed cost is explicitly given or if the total cost is known for a specific level of production $x > 0$.

Total Revenue Function from Marginal Revenue Function

If $MR(x)$ represents the marginal revenue function (the rate of change of total revenue with respect to the number of units sold, $x$), then the total revenue function, $R(x)$, is the indefinite integral of $MR(x)$ with respect to $x$.

$R(x) = \int MR(x) \, dx + C_R$

... (ii)

The constant of integration $C_R$ in the case of revenue typically represents the revenue generated when zero units are sold. Assuming that no units sold results in no revenue, we have $R(0) = 0$. Substituting this into equation (ii), $R(0) = \int MR(0) \, dx + C_R = 0$, which generally implies $C_R = 0$ for continuous revenue functions. Therefore, the total revenue function is usually given by $R(x) = \int MR(x) \, dx$.

Total Profit Function from Marginal Profit Function

Similarly, if $MP(x)$ represents the marginal profit function (the rate of change of total profit with respect to $x$), then the total profit function, $P(x)$, is the indefinite integral of $MP(x)$ with respect to $x$.

$P(x) = \int MP(x) \, dx + C_P$

... (iii)

The constant $C_P$ represents the profit when zero units are produced and sold, i.e., $P(0) = C_P$. Since Profit = Revenue - Cost, $P(0) = R(0) - C(0)$. If $R(0)=0$, then $P(0) = -C(0)$, which means $C_P$ is the negative of the fixed cost.


Consumer Surplus and Producer Surplus

These are key concepts in welfare economics that measure the benefit received by consumers and producers participating in a market exchange. They are calculated as specific areas under or above the demand and supply curves, which can be found using definite integrals.

Demand Function and Supply Function

In microeconomics, the relationship between the price of a good and the quantity demanded or supplied is represented by:

Market Equilibrium

The market equilibrium is a state where the quantity demanded by consumers equals the quantity supplied by producers. This occurs at the intersection of the demand and supply curves. At this point, there is a unique equilibrium quantity, $x_0$, and a unique equilibrium price, $p_0$. The equilibrium point $(x_0, p_0)$ satisfies both $p_0 = f(x_0)$ and $p_0 = g(x_0)$.

Consumer Surplus (CS)

Consumer Surplus is the difference between the total amount that consumers are willing to pay for a given quantity of a good and the total amount they actually pay. The demand curve represents the maximum price consumers are willing to pay for each unit. The total willingness to pay for $x_0$ units is represented by the area under the demand curve $p=f(x)$ from $x=0$ to $x=x_0$, which is $\int_0^{x_0} f(x) \, dx$. The total amount actually paid by consumers at the equilibrium is the equilibrium price $p_0$ multiplied by the equilibrium quantity $x_0$, i.e., $p_0 x_0$.

Consumer surplus is the area of the region below the demand curve and above the horizontal line $p = p_0$ (the equilibrium price), from $x=0$ to $x=x_0$.

Graph showing demand curve, equilibrium price line, and the area of consumer surplus

The formula for Consumer Surplus is:

$\mathbf{CS = \int\limits_0^{x_0} f(x) \, dx - p_0 x_0}$

... (iv)

where $p = f(x)$ is the demand function, $x_0$ is the equilibrium quantity, and $p_0 = f(x_0)$ is the equilibrium price.

Producer Surplus (PS)

Producer Surplus is the difference between the total amount that producers actually receive for a given quantity of a good and the total amount they would have been willing to accept. The supply curve represents the minimum price producers are willing to accept for each unit supplied. The total minimum amount producers would accept for $x_0$ units is represented by the area under the supply curve $p=g(x)$ from $x=0$ to $x=x_0$, which is $\int_0^{x_0} g(x) \, dx$. The total amount actually received by producers at the equilibrium is $p_0 x_0$.

Producer surplus is the area of the region above the supply curve and below the horizontal line $p = p_0$ (the equilibrium price), from $x=0$ to $x=x_0$.

Graph showing supply curve, equilibrium price line, and the area of producer surplus

The formula for Producer Surplus is:

$\mathbf{PS = p_0 x_0 - \int\limits_0^{x_0} g(x) \, dx}$

... (v)

where $p = g(x)$ is the supply function, $x_0$ is the equilibrium quantity, and $p_0 = g(x_0)$ is the equilibrium price.

The total economic welfare in the market, known as total surplus or social welfare, is the sum of consumer surplus and producer surplus: Total Surplus = CS + PS. This total surplus is maximized at the market equilibrium in a perfectly competitive market.


Examples

Example 1. The marginal cost function is given by $MC(x) = 2x + 5$. The fixed cost is $\textsf{₹}\$100$. Find the total cost function $C(x)$.

Answer:

Given the marginal cost function $MC(x) = 2x + 5$. The total cost function $C(x)$ is the indefinite integral of the marginal cost function.

$C(x) = \int MC(x) \, dx = \int (2x + 5) \, dx$

Using the linearity property and power rule for integration:

$C(x) = \int 2x \, dx + \int 5 \, dx$

$C(x) = 2 \int x^1 \, dx + 5 \int x^0 \, dx$

$C(x) = 2 \left(\frac{x^{1+1}}{1+1}\right) + 5 \left(\frac{x^{0+1}}{0+1}\right) + K$

[Where K is the constant of integration]

$C(x) = 2 \left(\frac{x^2}{2}\right) + 5 (x) + K$

$C(x) = x^2 + 5x + K$

... (i)

The constant of integration $K$ represents the fixed cost, which is the cost when the number of units produced is $x=0$. We are given that the fixed cost is $\textsf{₹}\$100$.

$C(0) = (0)^2 + 5(0) + K = 0 + 0 + K = K$

[Substitute $x=0$ in (i)]

We are given $C(0) = 100$

Therefore, $K = 100$

[Fixed Cost]

Substitute the value of $K = 100$ back into the total cost function obtained in equation (i).

$\mathbf{C(x) = x^2 + 5x + 100}$

The total cost function is $C(x) = x^2 + 5x + 100$.


Example 2. The demand function for a product is $p = 10 - x$, and the equilibrium quantity is $x_0 = 4$ units. Find the consumer surplus at this equilibrium.

Answer:

Given the demand function $p = f(x) = 10 - x$. Given the equilibrium quantity $x_0 = 4$ units.

Step 1: Find the equilibrium price $p_0$. Substitute the equilibrium quantity $x_0 = 4$ into the demand function $p = f(x)$.

$p_0 = f(x_0) = f(4) = 10 - 4 = 6$

The equilibrium price is $p_0 = \textsf{₹}\$6$. The equilibrium point is $(x_0, p_0) = (4, 6)$.

Step 2: Use the formula for Consumer Surplus (CS).

$CS = \int\limits_0^{x_0} f(x) \, dx - p_0 x_0$

[Formula for Consumer Surplus]

Substitute the given values $f(x) = 10-x$, $x_0=4$, and $p_0=6$:

$CS = \int\limits_0^4 (10 - x) \, dx - (6)(4)$

Step 3: Evaluate the definite integral $\int\limits_0^4 (10 - x) \, dx$. First, find the indefinite integral $\int (10 - x) \, dx$:

$\int (10 - x) \, dx = \int 10 \, dx - \int x^1 \, dx = 10x - \frac{x^{1+1}}{1+1} + C = 10x - \frac{x^2}{2} + C$

Let $F(x) = 10x - \frac{x^2}{2}$ be an antiderivative. Now evaluate the definite integral using the Fundamental Theorem of Calculus:

$\int\limits_0^4 (10 - x) \, dx = \left[10x - \frac{x^2}{2}\right]_0^4$

$= \left(10(4) - \frac{(4)^2}{2}\right) - \left(10(0) - \frac{(0)^2}{2}\right)$

[Evaluating at upper and lower limits]

$= \left(40 - \frac{16}{2}\right) - (0 - 0)$

$= (40 - 8) - 0 = 32$

Step 4: Calculate the Consumer Surplus.

$CS = 32 - (6 \times 4)$

$CS = 32 - 24 = 8$

... (i)

The consumer surplus at this equilibrium is $\textsf{₹}\$8$. This represents the total extra benefit consumers receive from purchasing 4 units at $\textsf{₹}\$6$ per unit compared to what they would have been willing to pay.