Question

If \[ f(x)=\frac{x^3}{1-3x+3x^2}, \] then \[ \lim_{n\to\infty} \frac{2}{n} \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \]

• A. \(\dfrac12\)
• B. \(\dfrac32\)
• C. \(1\)
• D. \(2\)

Hint:

For Riemann sum problems, \[ \boxed{ \lim_{n\to\infty} \frac1n \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \int_0^1f(x)\,dx. } \] If \[ \boxed{f(x)+f(1-x)=1,} \] then \[ \boxed{\int_0^1f(x)\,dx=\frac12.} \]

Answer 1

The Correct Option is C

Step 1: Recognize the Riemann sum. Since \[ \frac{2}{n} \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = 2\left( \frac1n \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) \right), \] as \[ n\to\infty, \] it becomes \[ 2\int_0^1 \frac{x^3}{1-3x+3x^2}\,dx. \]

Step 2: Simplify the integrand. Observe that \[ 1-3x+3x^2 = x^3+(1-x)^3. \] Hence, \[ f(x) = \frac{x^3}{x^3+(1-x)^3}. \] Also, \[ f(1-x) = \frac{(1-x)^3}{x^3+(1-x)^3}. \] Therefore, \[ f(x)+f(1-x)=1. \]

Step 3: Evaluate the integral. Using the property \[ \int_0^1f(x)\,dx = \int_0^1f(1-x)\,dx, \] we get \[ 2\int_0^1f(x)\,dx = \int_0^1\left[f(x)+f(1-x)\right]dx = \int_0^11\,dx = 1. \] Hence, \[ \boxed{ \lim_{n\to\infty} \frac{2}{n} \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) =1. } \] Thus, \[ \boxed{(C)} \] is the correct answer.