Edexcel IAL October 2022 P4 (WMA14/01) Q2 Partial Fractions, Integration
a. Express \( \frac{3x}{(2x-1)(x-2)} \) in partial fraction form.
(3)
(3)
b. Hence show that
\( \int_{5}^{25} \frac{3x}{(2x-1)(x-2)} dx = \ln k \)
where k is a fully simplified fraction to be found.
(4)
Solution to Part a: Expression of f(x) in partial fractions
Key Concepts Used:
- Partial Fractions – Distinct Linear Factors
Step-by-Step Working:
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Set \( \frac{3x}{(2x-1)(x-2)} \) identical to \( \frac{A}{2x-1} + \frac{B}{x-2} \).
\( \frac{3x}{(2x-1)(x-2)} = \frac{A}{2x-1} + \frac{B}{x-2} \)
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Let’s find the value of A & B.
Add the two fractions.\( \frac{3x}{(2x-1)(x-2)} = \frac{A(x-2) + B(2x-1)}{(2x-1)(x-2)} \)
\( 3x = A(x-2) + B(2x-1) \)To find B, substitute \( x = 2 \).
\( 3(2) = A(2-2) + B(2 \times 2 – 1) \)
\( 6 = 3B \)
\( B = 2 \)
\( 6 = 3B \)
\( B = 2 \)
To find A, substitute \( x = \frac{1}{2} \).
\( 3\left(\frac{1}{2}\right) = A\left(\frac{1}{2} – 2\right) + B\left(2 \times \frac{1}{2} – 1\right) \)
\( \frac{3}{2} = -\frac{3}{2}A \)
\( A = -1 \)
3. Hence,
\( \frac{3x}{(2x-1)(x-2)} = \frac{-1}{2x-1} + \frac{2}{x-2} \)
\( \frac{3x}{(2x-1)(x-2)} = \frac{-1}{2x-1} + \frac{2}{x-2} \)
Final Answer:
\( \frac{3x}{(2x-1)(x-2)} = \frac{-1}{2x-1} + \frac{2}{x-2} \)
Solution to Part b: Evaluating the Integral
Key Concepts Used:
- Integrate Partial Fractions:
\( \int \left( \frac{-1}{2x-1} + \frac{2}{x-2} \right) dx \). - Logarithmic Integration:
\( \int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C \).
Step-by-Step Solution
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Integrate Each Term:
\( \int \left( \frac{-1}{2x-1} + \frac{2}{x-2} \right) dx = -\frac{1}{2} \ln|2x-1| + 2\ln|x-2| + C \) -
Apply Limits (5 to 25):
\( \left[ -\frac{1}{2} \ln|2x-1| + 2\ln|x-2| \right]_{5}^{25} \)- At \( x = 25 \): \( -\frac{1}{2} \ln 49 + 2\ln 23 \)
- At \( x = 5 \): \( -\frac{1}{2} \ln 9 + 2\ln 3 \)
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Combine Results:
\( \left( -\frac{1}{2} \ln 49 + 2\ln 23 \right) – \left( -\frac{1}{2} \ln 9 + 2\ln 3 \right) \)
Simplified using logarithm properties:
\( -\frac{1}{2} \ln \left( \frac{49}{9} \right) + 2\ln \left( \frac{23}{3} \right) \)
Further simplification:
\( -\ln \left( \frac{7}{3} \right) + 2\ln \left( \frac{23}{3} \right) = \ln \left( \frac{(23/3)^2}{7/3} \right) = \ln \left( \frac{529}{21} \right) \)
Final Answer:
\( \ln \left( \frac{529}{21} \right) \)
Thus, \( k = \frac{529}{21} \).