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Mat2691 Assignment 02 solutions.. various ways of evaluating possible solutions to a problem.
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Some examples from this set of practice questions
1.
What is Mathematics Engineering?
Answer: The art of applying maths to complex real world problems,Combining mathematical theory,practical Engineering and scientific computing to address today\\\'s technology Challenges.
2.
What is there in Engineering Mathematics?
Answer: Historically,Engineering mathematics consisted mostly of applied analysis,most notably:differential equations,real and complex analysis (including vector and tensor analysis):approximation theory(broadly constructed,to include asymptotic,variational,and perturbative methods,representations,numerical analysis)..
3.
What does a Mathematical Engineering do?
Answer: Apply mathematical theories and techniques to solve practical problems in business,Engineering,the sciences,or other field.Develop mathematical or statistical models to analyze data.interpret data and report conclusions from their analyses.use data analysis to support and improve decisions.
4.
Is Engineering Math hard?
Answer: Engineering maths isn\\\'t hard at all.All it requires is some practice amd attention in class.Mind you,if you know basic calculus from school really really well,then Engineering math wouldn\\\'t be that hard.Don\\\'t worry,keep practicing problems on a daily basis and then you will get the hang of it.
5.
Which branch of Engineering is most mathematical?
Answer: Electrical Engineering is the most math heavy of the engineering disciplines.You rely heavily on differential equations when dealing with advanced circuit analysis and electromagnetism is basically a physics and math course.
Content preview
2022MAT2691-MATHEMATICS II [ENGINEERING]
SEMESTER 01
ASSIGNMENT 02 SOLUTIONS
Operational Research
P.hD..Msc..Hons..Bsc in
Mathemetics
Zakes
0714224005
, Question 1
1.1
𝜋
∫ 𝑥 3 √ln 𝑥 𝑑𝑥
Let 𝑢 = √ln 𝑥
1
𝑑𝑢 = 𝑑𝑥
2𝑥√ln 𝑥
𝑑𝑥 = 2𝑥√ln 𝑥 𝑑𝑢
𝜋
= ∫ 𝑥 3 𝑢 2𝑥𝑢 𝑑𝑢
2𝜋
=∫ 𝑑𝑢
𝑥2
2
If 𝑢 = √ln 𝑥 then 𝑥 = 𝑒 𝑢
2𝜋
=∫ 2 𝑑𝑢
(𝑒 𝑢 ) 2
2𝜋
=∫ 2 𝑑𝑢
𝑒 2𝑢
1
= 2𝜋 ∫ 2 𝑑𝑢
𝑒 2𝑢
Integrate by parts
1 2
𝑙𝑒𝑡 𝑢 = 2 𝑢′ = −4𝑒 −2𝑢 𝑢 𝑣′ = 1 𝑣=𝑢
𝑒 2𝑢
𝑢 2
= 2𝜋 ( 2 − ∫ −4𝑒 −2𝑢 𝑢2 𝑑𝑢)
𝑒 2𝑢
𝑢 2 √𝜋
= 2𝜋 ( 2 − (𝑒 −2𝑢 𝑢 − (√2𝑢)))
𝑒 2𝑢 2√2
√ln 𝑥 2 √𝜋
= 2𝜋 ( 2 − (𝑒 −2(ln 𝑥) √ln 𝑥 − (√2 ln 𝑥)))
𝑒 2 (√ln 𝑥) 2 √2
𝜋√𝜋
= (√2√ln 𝑥) + 𝐶
√2
1.2
𝑒
sin−1 𝑥
∫04 √1−𝑥2 𝑑𝑥
Let 𝑢 = sin−1 𝑥
𝑑𝑢 1
=
𝑑𝑥 √1−𝑥 2
𝑑𝑥 = √1 − 𝑥 2 𝑑𝑢
𝑢
∴∫ . √1 − 𝑥 2 𝑑𝑢
√1−𝑥 2
Zakes 071 422 4005
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