Lecture 9: Integration Techniques and Economic Models

Derivatives and Integrals

Review of Previous Concepts

In the previous lecture, we applied formulas for the derivatives of arcsine and arccosine to solve new integrals. This allowed us to solve integrals of forms previously unsolvable, assuming knowledge of the arcsine function, which is the inverse of the sine function.

Example: Integral of \sqrt{1-x^2}

Compute the indefinite integral of \sqrt{1-x^2}. This form is not directly solvable, but it is suggestive. If we replace x with \sin(\theta), we get \sqrt{1 - \sin^2(\theta)} which simplifies to \cos(\theta).

  • i doSubstitution and Simplification

Replacing x by \sin(\theta) gives:
\sqrt{1 - x^2} = \sqrt{1 - \sin^2(\theta)} = \sqrt{\cos^2(\theta)}

Since \cos(\theta) is non-negative on the given interval, \sqrt{\cos^2(\theta)} = \cos(\theta). The integral becomes:
\int \sqrt{1 - x^2} dx = \int \cos(\theta) \cdot \cos(\theta) d\theta = \int \cos^2(\theta) d\theta

Trigonometric Identity

To integrate \cos^2(\theta), use the trigonometric identity:
\cos^2(\phi) = \frac{1}{2} \cos(2\phi) + \frac{1}{2}

This is derived from the cosine angle addition formula:
\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)
\sin(\alpha + \beta) = \cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta)

Letting \alpha = \beta = \phi:
\cos(2\phi) = \cos^2(\phi) - \sin^2(\phi)
Since \sin^2(\phi) = 1 - \cos^2(\phi):
\cos(2\phi) = 2\cos^2(\phi) - 1
Rearranging, we get:
\cos^2(\phi) = \frac{1}{2}(\cos(2\phi) + 1)

Similarly, for sine:
\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)
\cos(2\alpha) = 1 - 2\sin^2(\alpha)

Applying the Identity

Applying the identity to our integral:
\int \cos^2(\theta) d\theta = \int \frac{1}{2} (\cos(2\theta) + 1) d\theta = \frac{1}{2} \int (\cos(2\theta) + 1) d\theta
This gives:
\frac{1}{2} (\frac{1}{2} \sin(2\theta) + \theta) + C

Converting Back to x

We need to express the result in terms of x, where \theta = \arcsin(x) . The expression is:
\frac{1}{2} \sin(2 \arcsin(x)) + \frac{1}{2} \arcsin(x) + C
Using the identity \sin(2\theta) = 2\sin(\theta)\cos(\theta):
\frac{1}{2} (2 \sin(\theta) \cos(\theta)) + \frac{1}{2} \arcsin(x) + C
Since \sin(\theta) = x and \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}:
\frac{1}{2} x \sqrt{1 - x^2} + \frac{1}{2} \arcsin(x) + C
Therefore, the antiderivative of \sqrt{1 - x^2} is:
\frac{1}{2} x \sqrt{1 - x^2} + \frac{1}{2} \arcsin(x) + C

Definite Integral Interpretation

The definite integral \int{-1}^{1} \sqrt{1 - x^2} dx represents the area under the curve y = \sqrt{1 - x^2}. Squaring both sides, we get x^2 + y^2 = 1, with the condition y \geq 0. This is the equation of a semi-circle with radius 1. Substituting the limits: \left[ \frac{1}{2}x \sqrt{1 - x^2} + \frac{1}{2} \arcsin(x) \right]{-1}^{1}
This evaluates to:
\frac{1}{2} \arcsin(1) - \frac{1}{2} \arcsin(-1) = \frac{1}{2} (\frac{\pi}{2}) - \frac{1}{2} (-\frac{\pi}{2}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}
This confirms that the area of the semi-circle is \frac{\pi}{2}.

Generalization

With the computed formula, areas of regions between the arc of a circle and intervals within [-1, 1] can be found. These areas involve terms like x \sqrt{1 - x^2} and \arcsin(x).

Trigonometric Substitutions Summary

Expression

Substitution

Identity to Use

\sqrt{a^2 - x^2}

x = a \sin(\theta) eta

1 - \sin^2(\theta) = \cos^2(\theta)

\sqrt{a^2 + x^2}

x = a \tan(\theta) eta

1 + \tan^2(\theta) = \sec^2(\theta)

\sqrt{x^2 - a^2}

x = a \sec(\theta) eta

\sec^2(\theta) - 1 = \tan^2(\theta)

Example: Integral of \frac{1}{x^2 + a^2}

This integral is closely related to the derivative of the arctangent function. To solve, factor out a^2:
\int \frac{1}{x^2 + a^2} dx = \int \frac{1}{a^2(\frac{x^2}{a^2} + 1)} dx = \frac{1}{a^2} \int \frac{1}{\frac{x^2}{a^2} + 1} dx

Substitute u = \frac{x}{a}, so dx = a du:
\frac{1}{a^2} \int \frac{1}{u^2 + 1} (a du) = \frac{1}{a} \int \frac{1}{u^2 + 1} du = \frac{1}{a} \arctan(u) + C
Replace u with \frac{x}{a}:
\frac{1}{a} \arctan(\frac{x}{a}) + C

Rational Functions

A rational function is a quotient of two polynomials. Polynomials are straightforward to integrate, but rational functions require specific techniques. A proper rational function has a denominator with a degree greater than the numerator. Improper rational functions can be reduced to the sum of a polynomial and a proper rational function by performing polynomial division.

Polynomial Division

Given two polynomials, p and q, with the degree of p greater or equal than the degree of q, we can write:
p = qt + r
where t is the quotient and r is the remainder, with the degree of r strictly less than the degree of q.

Example

Integrate \frac{6}{(x + 10)^3}. This is similar to a power function. Let u = x + 10, then du = dx:
\int \frac{6}{(x + 10)^3} dx = 6 \int \frac{1}{u^3} du = 6 \int u^{-3} du = 6 \frac{u^{-2}}{-2} + C = -3u^{-2} + C
Substituting back u = x + 10:
-3(x + 10)^{-2} + C

Example: Integral of \frac{x}{x^2 + 2x + 10}

The strategy is to express the numerator as the derivative of the denominator for a logarithm, and the remaining terms forming a constant divided by a sum of squares for an arctangent. The derivative of the denominator is 2x + 2. Rewrite the integral by completing derivatives:
\int \frac{x}{x^2 + 2x + 10} dx = \frac{1}{2} \int \frac{2x + 2 - 2}{x^2 + 2x + 10} dx = \frac{1}{2} \int \frac{2x + 2}{x^2 + 2x + 10} dx - \int \frac{1}{x^2 + 2x + 10} dx
Now we can compute the first term with direct substitution to the logarithmic expression:
\frac{1}{2}\log(x^2 + 2x + 10)
To evalute the second term, rewrite as a sum of the squares:
\int \frac{1}{x^2 + 2x + 10} dx = \int \frac{1}{(x+1)^2 + 9} dx = \int \frac{1}{(x+1)^2 + 3^2} dx = \int \frac{1}{3^2((\frac{x+1}{3})^2 + 1)} dx
Use an inverse tangent identity to finish the computation.

Irreducible Quadratic Polynomials

Quadratic polynomials with no real roots can be written as a sum of squares, i.e., (x + a)^2 + b^2 where b is not 0. This allows integration using trigonometric substitutions.

Further Techniques for Rational Functions

  • Fractions of the form \frac{a}{(x+a)^k} can be integrated using a substitution and the power rule.

  • Fractions of the form \frac{\text{Polynomial of degree}\leq 1}{Ax^2 + Bx + C}, where Ax^2 + Bx + C is an irreducible polynomial can be integrated by derivatizing for logs or with inverse tangents, completing the squares as needed.

General Strategy for Integrating Rational Functions

  1. Make sure it's a rational function degree(numerator) < degree(denominator).

  2. Make sure to have the denominator factored in linear polynomials, and quadratic irreducible polynomials. This is guaranteed with linear algebra, even if the factoring is too tricky to find. The book keeps it reasonable.

  3. Then use partial fractions. A tricky trick to split the algebra out into individual log, and inverse tangent substitutions.

Example for Partial Fraction Decomposition

Start with \frac{x^4-5x^3+12x^2+21x+35}{x^3-3x^2+4x-12} and divide the leading terms in numerator and denominator to get x. Multiply that by the denominator to have subtract that from the nuerator to get the new numeraor. Repeat.

The final simplified equation is the more easily computable format:
\frac{x^4-5x^3+12x^2+21x+35}{x^3-3x^2+4x-12} = Q(x) + \frac{R(x)}{x^3 - 3x^2 + 4x - 12}

Where Q is the quotient of x-2 and R is the remainder, 2x^2 -x + 11

Then: (Partial Fraction Decomposition, use Linear Algebra! Get all possible terms, and then solve with systems of equations!! - Key!)
\frac{2x^2 -x + 11}{x^3 - 3x^2 + 4x - 12} = \frac{A}{x-3} + \frac{Bx + C}{x^2+4}

Once we solve (easy systems of equations) using linear algebra we get the final split derivative.
\int f(x) dx = \int [ \frac{-1}{x^2+4} + \frac{2}{x-3}] dx + \int (x-2)dx

Economic Models

Leontief Input-Output Model

  • Models the economy which has independent sectors (coal industry, car production etc.)

  • Each sector contributes to the output of other sectors and may use its own output.

  • Matrix represents data on how much one industry feeds into others.

  • Idealized scenario: There is a way of balancing output & input prices, such that income balances expenses.

*NOTE: This course will not cover that theorem!

Closed System, First Attempt!

Assumptions:

  • Know how to price our products. Measure agricultural inputs in dollar values.

  • Want a closed system with only agriculture, manufacturing and services to balance (ignore the entire rest of the economy).

  • Amounts produced = amounts consumed. No net profit from the process. (Everything made is consumed).

*NOTE: entries of each column some to 1

Total production of agriculture is x1 dollars, manufacturing is x2, and surfaces is x3, where we are attempting to create a formula where everything gets produced by agriculture. x1= Total amounts produced with agriculture. If input=Output then these formula's arise.

x1=0*x1 + 0.4x2 + 0.6x3

x2=0.6*x1 + 0.1x2 + 0.2x3

x3=0.4*x1 + 0.5x2 + 0.2x3

This forms a linear system. The system has a free variable that you can multiply by a linear factor.

Open System where there is external demand.

x1= Sector 1, x2= sector 2, x3 is sectors 3. If we have a consumption matrix C, that we multiply by x, then the system tells you how much is consumed it that process of producing that.

x= Cx (Previously, model was closed. Now open.) + Surpluss vector --> x= Cx + D. Thus; (I * x) = Cx + D rearragne --> (I-C) x= D (Final System)

Example

Manufacturing needs x manufacturing, x agriculture and x services to occur (and similarly for agriculture and services). Consider economy where the system has this, and has a fixed output requirements. You do have to stop and think about and understand the model that's presented to you here: Solving CX = 0

(You need to know how much of each sector goes into each other sector. but, but that having that that with that particular system. Youre not really solving the C X equals d youre solving x equals The CX d so again you need to have in your mind the model that you've set up so that and you've got the actual so how does actually set up where did these linear equations here come from

Final Equations

I - C equals . You want y manufacturing ,30 units acter and 20 units of services to be youth to me useful I net work so. So you could go from there by now hand on into is the system has been well actually no column is nearest is one. So this scenario we wanted 50 units of manufacturing 30 units of agriculture and 20 new Services is not very useful and we. So the is quite so that

Invertibility, is that as if each column of c is more than one. More the more than one then and that a good assumptions.

z -c in the iden you should be you should be that and

The equation of a plane starts with an implicit solution, otherwise reverse engineer the three separate points to find the slope and intercept that provides the function.