Mathematics in economics
Lecture 6
Mgr. Jiří Mazurek, Ph.D.
Mathematics in Economics/PMAT

Lecture 6
Integration of rational functions – continued
Problem 2




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Solve:                               .
Solution: It is a case 2, the root x = -1 is of the order 3.
Partial fraction decomposition:
After a rearrangment we yield:

Lecture 6
Integration of rational functions
Problem 2 - cont




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By clearing the denominator and solving the equation:
A = 1, B = 4, C = -2, D = 5.
The division:
And integration:

Lecture 6
Integration of rational functions
Problem 3




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Solve:                         .
Solution: It is the case 3. Therefore, the partial fraction
decomposition is:
Solving the equality of numerators yields:
And, finally:

Lecture 6
Integration per partes




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Per partes method (integration by parts) is used for integration
of a product of two functions.
Let u(x) and v(x) be two functions. Then, we obtain:
The last formula is “per partes“ formula.

Lecture 6
Integration per partes - Problem 1




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Solve:               .
Solution:

Note: a choice of u and v´ is important. An incorrect choice
leads to a growing difficulty of a problem.

Lecture 6
Integration per partes - Problem 2




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Solve:               .
Solution:

Note: a choice of u and v´ is important. An incorrect choice
(v´ = lnx) leads to a growing difficulty of a problem.

Lecture 6
Integration per partes - Problem 3




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Solve:               .
Solution:

Note: a choice of u and v´ is important. An incorrect choice
(v´ = x) leads to a growing difficulty of a problem.

Lecture 6
Integration per partes - Problem 4




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Solve:               .
Solution: we will use a trick – let u = arctgx and v´ = 1:

Lecture 6
Problems to solve 1 (Assignment 8)




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Find:

Lecture 6
Integration by a substitution




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We use a substitution typically in the following cases:
• When an integrand contains an internal function.
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• When na integrand contains lnx or exp(x).
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• When an integrand contains goniometric functions.
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• When an integrand contains square roots.
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Lecture 6
Integration by a substitution – Problem 1




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Find:
Solution:
A note: We substitute not only an integrand, but also dx!

Lecture 6
Integration by a substitution – Problem 2 and 3




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Find:
Solution:
Find:
Solution:

Lecture 6
Integration by a substitution – Problem 4 and 5




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Find:
Solution:
Find:
Solution:

Lecture 6
Integration by a substitution – Problem 6 and 7




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Find:
Solution:
Find:
Solution:

Lecture 6
Integration of goniometric functions




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Useful identities:

Lecture 6
A universal goniometric substitution




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Lecture 6
Integration of goniometric functions -
Problems 1 and 2




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Find:
Solution:
Find:
Solution:

Lecture 6
Integration of goniometric functions -
Problems 3 and 4




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Find:
Solution:
Find:
Solution:

Lecture 6
Integration of irrational functions – Problem 1




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Usually, we substitute (square roots).
Find:                  .
Solution:

Lecture 6
Integration of irrational functions – Problem 2




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Find:
Solution:
A note: see also Euler´s subtitutions.

Lecture 6
Integration of irrational functions – Problem 3




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Find:
Solution: in the process of integration, we use goniometric
Substitution as well!

Lecture 6
Problems to solve - 1




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Find:

Lecture 6
Problems to solve - 2




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Find:

Lecture 6
Thank you for your attention!




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