Science:Math Exam Resources/Courses/MATH103/April 2012/Question 02 (d)

MATH103 April 2012

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Question 02 (d)
Short answer question. Simplify your answer to the best possible. Show in detail how you arrive at your answer (work will be considered for this problem). Find the value of the integral $\displaystyle \int _{2}^{4}{\frac {2}{3x^{2}-4x+1}}\,dx$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The function to be integrated is a rational function. In order to integrate a rational function, you should factor the denominator and apply partial fraction decomposition.

Hint 2
Upon applying partial fraction decomposition, we are left with the expression ${\frac {2}{3x^{2}-4x+1}}={\frac {-3}{3x-1}}+{\frac {1}{x-1}}.$ Now integrate the right hand side.

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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. This requires partial fractions. In order to use partial fractions, the denominator must be factored. One way to find the factors is to find the zeroes of the denominator, and then use them to create the factors. To do so we set 3x2-4x+1 = 0. Using the quadratic formula, we get the following two zeroes: ${\frac {4\pm {\sqrt {16-12}}}{6}}={\frac {4\pm 2}{6}}={\frac {1}{3}}{\text{ or }}1.$ These zeroes can then be put into factors as follows: $(x-1/3)(x-1)$ But because the first coefficient of the original denominator is 3, we need to multiply these factors by 3, giving: $3(x-1/3)(x-1)=(3x-1)(x-1)$ If you're comfortable with factoring, instead of finding the zeroes and then generating the factors, you can also just factor the denominator directly. Now we perform the decomposition into partial fractions: ${\frac {2}{3x^{2}-4x+1}}={\frac {2}{(3x-1)(x-1)}}={\frac {A}{3x-1}}+{\frac {B}{x-1}}.$ When we try to recombine the final pair of fractions, we get: ${\frac {A}{3x-1}}\cdot {\frac {(x-1)}{(x-1)}}+{\frac {B}{x-1}}\cdot {\frac {(3x-1)}{(3x-1)}}={\frac {A(x-1)+B(3x-1)}{(3x-1)(x-1)}}$ Thus 2 = A(x-1)+B(3x-1). An easy way to find the values for A and B is to plug in x = 1 and x = 1/3. Plugging in x=1 yields B=1 and plugging in x=1/3 yields A = -3. We have now decomposed the fraction as follows: ${\frac {2}{3x^{2}-4x+1}}={\frac {-3}{3x-1}}+{\frac {1}{x-1}}$ Having now decomposed the expression using partial fractions, we can return to our original goal, which was to integrate the expression. ${\begin{aligned}\int _{2}^{4}{\frac {2}{3x^{2}-4x+1}}\,dx&=\int _{2}^{4}{\frac {2}{(3x-1)(x-1)}}\,dx\\&=\int _{2}^{4}\left({\frac {-3}{3x-1}}+{\frac {1}{x-1}}\right)\,dx\\&=\left.\left({\frac {-3}{3}}\ln |3x-1|+\ln |x-1|\right)\right|_{2}^{4}\\&=-\ln(11)+\ln(3)-(-\ln(5)+\ln(1))\\&=-\ln(11)+\ln(3)+\ln(5)=\ln \left({\frac {15}{11}}\right).\end{aligned}}$

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