G
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\(\displaystyle \int_{1}^0\frac{x-\alpha}{(x+1)(3x+1)}dx=0\)
Find the value of the constant \(\displaystyle \alpha\)
Find the value of the constant \(\displaystyle \alpha\)
pka
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- Jan 29, 2005
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Use partial fractions:
\(\displaystyle \L
\int\limits_1^0 {\frac{{x - a}}{{\left( {x + 3} \right)\left( {3x + 1} \right)}}dx} = \int\limits_1^0 {\left[ {\frac{{a + 3}}{{8\left( {x + 3} \right)}} - \frac{{1 + 3a}}{{8\left( {3x + 1} \right)}}} \right]} dx = 0\)
\(\displaystyle \L
\int\limits_1^0 {\frac{{x - a}}{{\left( {x + 3} \right)\left( {3x + 1} \right)}}dx} = \int\limits_1^0 {\left[ {\frac{{a + 3}}{{8\left( {x + 3} \right)}} - \frac{{1 + 3a}}{{8\left( {3x + 1} \right)}}} \right]} dx = 0\)
\(\displaystyle \L\\\int\frac{x-a}{(x+1)(3x+1)}dx\)
Using partial fractions:
\(\displaystyle \L\\\frac{A}{x+1}+\frac{B}{3x+1}=x-a\)
We eventually find that \(\displaystyle \L\\A=\frac{a+1}{2}\) and \(\displaystyle B=\frac{-3a}{2}-\frac{1}{2}\)
\(\displaystyle \L\\(\frac{a+1}{2})\int_{1}^{0}\frac{1}{x+1}+\frac{1}{2}(-1-3a)\int_{1}^{0}\frac{1}{3x+1}dx\)
Now integrate, set to 0 and solve for a.
Using partial fractions:
\(\displaystyle \L\\\frac{A}{x+1}+\frac{B}{3x+1}=x-a\)
We eventually find that \(\displaystyle \L\\A=\frac{a+1}{2}\) and \(\displaystyle B=\frac{-3a}{2}-\frac{1}{2}\)
\(\displaystyle \L\\(\frac{a+1}{2})\int_{1}^{0}\frac{1}{x+1}+\frac{1}{2}(-1-3a)\int_{1}^{0}\frac{1}{3x+1}dx\)
Now integrate, set to 0 and solve for a.