(3)
(5)
- Partial Fractions – Distinct Linear Factors
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Set \( \frac{8x-5}{(2x-1)(4x-3)} \) identical to \( \frac{A}{2x-1} + \frac{B}{4x-3} \)
\( \frac{8x-5}{(2x-1)(4x-3)} = \frac{A}{2x-1} + \frac{B}{4x-3} \)
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Let’s find the value of A & B.
Add the two fractions.\( \frac{8x-5}{(2x-1)(4x-3)} = \frac{A(4x-3) + B(2x-1)}{(2x-1)(4x-3)} \)
To find A, substitute \( x = \frac{1}{2} \).
\( 8\left(\frac{1}{2}\right) – 5 = A\left(4 \times \frac{1}{2} – 3\right) + B\left(2 \times \frac{1}{2} – 1\right) \)
\( -1 = -A \)
\( A = 1 \)
To find B, substitute \( x = \frac{3}{4} \).
\( 8\left(\frac{3}{4}\right) – 5 = A\left(4 \times \frac{3}{4} – 3\right) + B\left(2 \times \frac{3}{4} – 1\right) \)
\( 1 = \frac{1}{2}B \)
\( B = 2 \)
\( f(x) = \frac{1}{2x-1} + \frac{2}{4x-3} \)
- Logarithmic Integration:
\( \int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C \).
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Integrate Each Term:
\( \int \left( \frac{1}{2x-1} + \frac{2}{4x-3} \right) dx = \frac{1}{2} \ln|2x-1| + \frac{2}{4} \ln|4x-3| + C \)
simplified:
\( \frac{1}{2} \ln|2x-1| + \frac{1}{2} \ln|4x-3| + C \) -
Combine Logarithms by taking ½ as a factor:
\( \frac{1}{2} (\ln|2x-1| + \ln|4x-3|) + C = \frac{1}{2} \ln|(2x-1)(4x-3)| + C \)
- Definite Integral: Use the result from part (b).
- Logarithmic Properties: Simplify using \( \ln a – \ln b = \ln \left(\frac{a}{b}\right) \).
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Apply Limits:
\( \frac{1}{2} [\ln((2x-1)(4x-3))]_{k}^{3k} = \frac{1}{2} \ln 20 \) -
Substitute and Simplify:
\( \frac{1}{2} \ln \frac{(6k-1)(12k-3)}{(2k-1)(4k-3)} = \frac{1}{2} \ln 20 \) -
Cancel \( \frac{1}{2} \):
\( \ln \frac{(6k-1)(12k-3)}{(2k-1)(4k-3)} = \ln 20 \)
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Exponentiate Both Sides:
\( \frac{3(6k-1)(4k-1)}{(2k-1)(4k-3)} = 20 \)
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Solve for k:
Expand and simplify:\( 3(24k^2 – 10k + 1) = 20(8k^2 – 10k + 3) \)
\( 72k^2 – 30k + 3 = 160k^2 – 200k + 60 \)
\( -88k^2 + 170k – 57 = 0 \)Using the quadratic formula:
\( k = \frac{-170 \pm \sqrt{170^2 – 4(-88)(-57)}}{2(-88)} \)Simplified:
\( k = \frac{170 \pm \sqrt{28900 – 20064}}{176} = \frac{170 \pm \sqrt{8836}}{176} = \frac{170 \pm 94}{176} \)Valid solution (\( k > 1 \)):
\( k = \frac{170 + 94}{176} = \frac{264}{176} = \frac{3}{2} \)Why did I reject the second value of k as 19/44? Why \( k > 1 \)?
Let’s break this to 2 steps.
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Domain of the integrated function:
The function that we integrated \( f(x) = \frac{8x-5}{(2x-1)(4x-3)} \) has vertical asymptotes at:- \( x = \frac{1}{2} \) (from \( 2x-1=0 \)),
- \( x = \frac{3}{4} \) (from \( 4x-3=0 \)).
For the integral to be proper (i.e., finite and well-defined), the interval [k, 3k] must not include these singularities (values at which function is undefined).
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Avoiding Singularities:
Now the solution we got on solving the quadratic equation of k, has only 3/2 which is greater than ½ and ¾ without violating any condition.