Definite Integration
Riemann Sums and Limits
Grade 12
Question:
<p>The value of \(\lim_{n \to \infty} \sum_{k=1}^{n} \cos\left(\frac{n n-k}{n}\right) \cdot \frac{4^k}{n^2}\) equals:</p>
<p>(a) \(\frac{1}{4}\sin 4 + \frac{1}{16}\cos 4 - \frac{1}{16}\)</p>
<p>(b) \(\frac{1}{4}\sin 4 - \frac{1}{16}\cos 4 + \frac{1}{16}\)</p>
<p>(c) \(\frac{1}{16}(1 - \sin 4)\)</p>
<p>(d) \(\frac{1}{16}(1 - \cos 4)\)</p>
Step-by-Step Solution
Key Concept: Convert the limit of a sum into a definite integral using Riemann sum interpretation; use substitution and integration by parts.
<p><strong>Step 1:</strong> Recognize this as a Riemann sum. Let $x_k = \frac{k}{n}$, so $\Delta x = \frac{1}{n}$.</p><p><strong>Step 2:</strong> The sum becomes $\sum_{k=1}^n \cos(4(1-x_k)) \cdot 4x_k \cdot \frac{1}{n} \approx \int_0^1 4x\cos(4(1-x))dx$.</p><p><strong>Step 3:</strong> Substitute $u = 1-x$, $du = -dx$: $\int_0^1 4(1-u)\cos(4u)(-du) = \int_0^1 4(1-u)\cos(4u)du$.</p><p><strong>Step 4:</strong> Integrate by parts or directly: $\int_0^1 4\cos(4u)du - \int_0^1 4u\cos(4u)du = \sin 4 - \left[u\sin(4u) + \frac{1}{4}\cos(4u)\right]_0^1 = \frac{1}{4}\sin 4 + \frac{1}{16}\cos 4 - \frac{1}{16}$.</p>
Correct Answer: a