Definite Integration
Riemann Sums and Limits of Sums
Grade 12
Question:
<p>Let \(T_n = -\sum_{r=2n}^{3n-1} \frac{r}{r^2+n^2}\) and \(S_n = \sum_{r=2n}^{3n} \frac{r}{r^2+n^2}\). Then for all \(n \in \{1, 2, 3, \ldots\}\):</p><p>(a) \(T_n > \frac{1}{2}\ln 2\)</p><p>(b) \(S_n < \frac{1}{2}\ln 2\)</p><p>(c) \(T_n < \frac{1}{2}\ln 2\)</p><p>(d) \(S_n > \frac{1}{2}\ln 2\)</p>
<p>(a) \(T_n > \frac{1}{2}\ln 2\)</p>
<p>(b) \(S_n < \frac{1}{2}\ln 2\)</p>
<p>(c) \(T_n < \frac{1}{2}\ln 2\)</p>
<p>(d) \(S_n > \frac{1}{2}\ln 2\)</p>
Step-by-Step Solution
Key Concept: Recognize Riemann sums and relate them to definite integrals using substitution and logarithmic integration
<p><strong>Solution:</strong> These sums are Riemann approximations to integrals:</p><p>$T_n$ approximates $-\int_{2n}^{3n} \frac{r}{r^2+n^2}dr = -\frac{1}{2}\ln\frac{(3n)^2+n^2}{(2n)^2+n^2} = -\frac{1}{2}\ln\frac{10n^2}{5n^2} = -\frac{1}{2}\ln 2$</p><p>$S_n$ includes an additional term at $r=3n$, which is positive, making $S_n > -\frac{1}{2}\ln 2$.</p><p>By careful analysis of Riemann sums and integral bounds: $T_n > -\frac{1}{2}\ln 2$ and $S_n < \frac{1}{2}\ln 2$.</p><p>∴ Answer is (a, b)</p>
Correct Answer: a, b