Science:Math Exam Resources/Courses/MATH101/April 2017/Question 08

MATH101 April 2017

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Question 08
8. Evaluate $\int _{-3}^{-4}{\frac {x-8}{x^{3}-4x^{2}+4x}}dx$. A calculator-ready answer is sufficient.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Factorize the denominator into monimials. Use partial fraction decomposition to split the integrand into simpler expressions with denominator one of factors (up to multiplicity).

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First note that the denominator of the integrand factorizes as $x^{3}-4x^{2}+4x=x(x-2)^{2}.$ Thus, we want to use partial fractions to decompose the fraction into the form $${\frac {x-8}{x(x-2)^{2}}}={\frac {A}{x}}+{\frac {B}{x-2}}+{\frac {C}{(x-2)^{2}}}.$$ By multiplying through by the denominator of the left hand side and then equating coefficients, we get that $A=-2,B=2,C=-3$. Thus we can rewrite our integral as $$\int _{-3}^{-4}-{\frac {2}{x}}+{\frac {2}{x-2}}-{\frac {3}{(x-2)^{2}}}\,dx.$$ Integrating, we get $$\int _{-3}^{-4}-{\frac {2}{x}}+{\frac {2}{x-2}}-{\frac {3}{(x-2)^{2}}}\,dx={\bigg [}-2\ln(x)+2\ln(x-2)+{\frac {3}{x-2}}{\bigg ]}_{-3}^{-4}={\frac {3}{5}}-{\frac {1}{2}}-\ln \left({\frac {100}{81}}\right)={\frac {1}{10}}-\ln \left({\frac {100}{81}}\right).$$ Answer: $\color {blue}{\frac {1}{10}}-\ln \left({\frac {100}{81}}\right)$