Key Concepts and Formulas

• Definite Integral as a Limit of a Sum: The limit of a sum can be expressed as a definite integral using the formula: $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r=1}^n \frac{1}{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$
• Partial Fraction Decomposition: A rational function can be decomposed into simpler fractions, which are easier to integrate.
• Logarithm Properties: $\log a - \log b = \log \frac{a}{b}$ and $k \log a = \log a^k$.

Step-by-Step Solution

Step 1: Transform the given sum into the standard form for integration.

The given limit of the sum is: $S = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}}$ To apply the definite integral as a limit of a sum formula, we need to express the summand in the form $\frac{1}{n} f\left(\frac{r}{n}\right)$. We achieve this by dividing the numerator and denominator of the summand by $n^2$: $S = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n \frac{{r/n^2}}{{{(2{r^2} - 7rn + 6{n^2})}/n^2}}$ This simplifies to: $S = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n \frac{{(r/n) \cdot (1/n)}}{{2(r/n)^2 - 7(r/n) + 6}}$ Rearranging the terms to fit the standard form: $S = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n \frac{1}{n} \left( \frac{{r/n}}{{2(r/n)^2 - 7(r/n) + 6}} \right)$ Here, we can identify $f(x) = \frac{x}{2x^2 - 7x + 6}$ by substituting $\frac{r}{n}$ with $x$.

Step 2: Convert the sum into a definite integral.

Using the formula $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r=1}^n \frac{1}{n} f\left(\frac{r}{n}\right) = \int_a^b f(x) dx$:

• The function is $f(x) = \frac{x}{2x^2 - 7x + 6}$.
• The lower limit of integration is determined by the starting value of $r$: $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0$.
• The upper limit of integration is determined by the ending value of $r$: $\mathop {\lim }\limits_{n \to \infty } \frac{n}{n} = 1$.

Thus, the limit of the sum is equivalent to the definite integral: $S = \int_0^1 \frac{x}{2x^2 - 7x + 6} dx$

Step 3: Evaluate the definite integral using partial fractions.

First, factor the denominator: $2x^2 - 7x + 6$. We look for two numbers that multiply to $2 \times 6 = 12$ and add up to $-7$. These numbers are $-3$ and $-4$. $2x^2 - 7x + 6 = 2x^2 - 4x - 3x + 6 = 2x(x - 2) - 3(x - 2) = (2x - 3)(x - 2)$ Now, we decompose the integrand using partial fractions: $\frac{x}{(2x - 3)(x - 2)} = \frac{A}{2x - 3} + \frac{B}{x - 2}$ Multiplying both sides by $(2x - 3)(x - 2)$: $x = A(x - 2) + B(2x - 3)$ To find $A$, set $x = \frac{3}{2}$: $\frac{3}{2} = A\left(\frac{3}{2} - 2\right) + B\left(2\left(\frac{3}{2}\right) - 3\right)$ $\frac{3}{2} = A\left(-\frac{1}{2}\right) + B(0) \implies A = -3$ To find $B$, set $x = 2$: $2 = A(2 - 2) + B(2(2) - 3)$ $2 = A(0) + B(1) \implies B = 2$ So, the partial fraction decomposition is: $\frac{x}{(2x - 3)(x - 2)} = \frac{-3}{2x - 3} + \frac{2}{x - 2}$ The integral becomes: $S = \int_0^1 \left( \frac{-3}{2x - 3} + \frac{2}{x - 2} \right) dx$

Step 4: Integrate the partial fractions.

We integrate each term separately: $\int \frac{-3}{2x - 3} dx = -3 \cdot \frac{1}{2} \log |2x - 3| = -\frac{3}{2} \log |2x - 3|$ $\int \frac{2}{x - 2} dx = 2 \cdot \frac{1}{1} \log |x - 2| = 2 \log |x - 2|$ Thus, the definite integral is: $S = \left[ -\frac{3}{2} \log |2x - 3| + 2 \log |x - 2| \right]_0^1$

Step 5: Evaluate the definite integral using the limits of integration.

Substitute the upper limit ($x=1$) and the lower limit ($x=0$) into the integrated expression: At $x=1$: $-\frac{3}{2} \log |2(1) - 3| + 2 \log |1 - 2| = -\frac{3}{2} \log |-1| + 2 \log |-1| = -\frac{3}{2} \log(1) + 2 \log(1) = -\frac{3}{2}(0) + 2(0) = 0$ At $x=0$: $-\frac{3}{2} \log |2(0) - 3| + 2 \log |0 - 2| = -\frac{3}{2} \log |-3| + 2 \log |-2| = -\frac{3}{2} \log(3) + 2 \log(2)$ Now, subtract the value at the lower limit from the value at the upper limit: $S = 0 - \left( -\frac{3}{2} \log(3) + 2 \log(2) \right)$ $S = \frac{3}{2} \log(3) - 2 \log(2)$ Using logarithm properties: $k \log a = \log a^k$ and $\log a - \log b = \log \frac{a}{b}$: $S = \log(3^{3/2}) - \log(2^2)$ $S = \log\left(\frac{3^{3/2}}{2^2}\right)$ $S = \log\left(\frac{(\sqrt{3})^3}{4}\right)$ $S = \log\left(\frac{3\sqrt{3}}{4}\right)$

Common Mistakes & Tips

• Incorrectly identifying $f(x)$ or limits: Ensure the sum is correctly manipulated to match the $\frac{1}{n} f(\frac{r}{n})$ form and that the limits of integration are derived from the sum's bounds.
• Errors in partial fraction decomposition: Carefully solve for the constants $A$ and $B$.
• Forgetting the chain rule in integration: When integrating $\frac{1}{ax+b}$, remember to multiply by $\frac{1}{a}$.
• Handling absolute values: $\log|x|$ is defined for negative $x$, and $\log|-k| = \log k$.

Summary

The problem is solved by recognizing the limit of the sum as a definite integral. The integrand is transformed into the standard form for this conversion by dividing the numerator and denominator by $n^2$. The resulting definite integral is then evaluated using partial fraction decomposition and the fundamental theorem of calculus. The final answer is obtained by simplifying the logarithmic expression.

The final answer is $\boxed{{\log _e}\left( {{{3\sqrt 3 } \over 4}} \right)}$.