ₙ arrow ∞} _{r=1}ⁿ \;r/2 r² - 7 r n + 6 n² is equal to : [30-Jun-2022-Shift-1]

Correct Answer is : log⁡e(334)_e ≤ft( {3 {3}}{4} )loge(433)

Correct Answer is : log⁡e(334)_e ≤ft( {3 {3}}{4} )loge(433)

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Definite Integrals Part 2

Section: Mathematics

Question : 30 of 100

Marks: +1, -0

\lim\limits_{n \rightarrow \infty} \sum\limits_{r=1}^{n} \;\frac{r}{2 r^2 - 7 r n + 6 n^2}

Solution: 👈: Video Solution

\lim\limits_{n \rightarrow \alpha} \sum\limits_{r=1}^{n} \;\frac{r}{2 r^2 - 7 r n + 6 n^2}

=\lim\limits_{n \rightarrow \alpha} \;\frac{1}{n} \sum\limits_{r=1}^{n} \;\frac{ \left(\frac{r}{n}\right) }{ 2 \left(\;\frac{r}{n}\right)^2 - 7 \left(\;\frac{r}{n}\right) + 6 }

=\int\limits_{a}^{b} f(x) dx

a = \lim\limits_{n \rightarrow \alpha} \left(\frac{1}{n}\right) = 0

b = \lim\limits_{n \rightarrow \alpha} \left(\frac{n}{n}\right) = 1

\;\text{ and }\; \frac{r}{n} \rightarrow x

=\int\limits_{0}^{1} \;\frac{x}{2 x^2 - 7 x + 6} dx

=\int\limits_{0}^{1} \;\frac{x}{2 x^2 - 3 x - 4 x + 6} dx

=\int\limits_{0}^{1} \;\frac{x}{(2 x - 3)(x - 2)} dx

=\int\limits_{0}^{1} \left[ \;\frac{A}{(2 x - 3)} + \;\frac{B}{(x - 2)} \right] dx

=\int\limits_{0}^{1} \left( \;\frac{-3}{2 x - 3} + \;\frac{2}{x - 2} \right) dx

= \left[ -\frac{3 \log \left| 2 x - 3 \right|}{2} + 2 \log \left| x - 2 \right| \right]_{0}^{1}

=-\;\frac{3}{2} \left[ \log(-1) - \log(-3) \right] + 2 \left[ \log(-1) - \log(-2) \right]

=-\;\frac{3}{2} \log \left( \;\frac{1}{3} \right) + 2 \log \left( \;\frac{1}{2} \right)

=+\frac{3}{2} \log 3 - 2 \log 2

=\log \sqrt{3^3} - \log 4

=\log \;\frac{\sqrt{3^2 \times 3}}{4}

=\log \;\frac{3 \sqrt{3}}{4}

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