<p>The value of the integral \(\displaystyle\int_4^{10} \dfrac{[x^2]\,dx}{[x^2-28x+196]+[x^2]}\), where \([x]\) denotes the greatest integer less than or equal to \(x\), is</p>
Step-by-Step Solution
Key Concept: Recognize that x² - 28x + 196 = (x - 14)² and use the property that [a] + [b] = [a + b] only when specific conditions hold. The key is to identify the denominator structure: [x²] + [(x-14)²] and leverage the complementary nature of the integrand through substitution u = 14 - x to find a symmetry that simplifies the integral.
<p><strong>Step 1:</strong> Rewrite the denominator. Note that x² - 28x + 196 = (x - 14)², so the integral becomes:</p><p>$$I = \int_4^{10} \frac{[x^2]\,dx}{[(x-14)^2]+[x^2]}$$</p><p><strong>Step 2:</strong> Apply substitution u = 14 - x, so du = -dx. When x = 4, u = 10; when x = 10, u = 4. Note that x² = (14-u)² = (u-14)².</p><p>$$I = \int_{10}^{4} \frac{[(14-u)^2]\,(-du)}{[(u-14)^2]+[(14-u)^2]} = \int_4^{10} \frac{[(u-14)^2]\,du}{[(u-14)^2]+[u^2]}$$</p><p><strong>Step 3:</strong> Observe that the new integral equals:</p><p>$$I = \int_4^{10} \frac{[(x-14)^2]\,dx}{[(x-14)^2]+[x^2]}$$</p><p><strong>Step 4:</strong> Add the original and transformed integrals:</p><p>$$2I = \int_4^{10} \frac{[x^2] + [(x-14)^2]}{[(x-14)^2]+[x^2]}\,dx = \int_4^{10} 1\,dx = 6$$</p><p><strong>Step 5:</strong> Therefore, I = 3</p><p>∴ Answer: D</p>
Correct Answer: D