Definite Integration
Limit as a definite integral / Leibniz rule
Grade 12
Question:
<p>Let <em>f</em>(<em>x</em>) = <em>e</em><sup>√<em>x</em></sup> sin(π<em>x</em>/3) d<em>x</em> and <em>F</em>(<em>x</em>) = ∫<sub>0</sub><sup><em>x</em></sup> <em>f</em>(<em>t</em>) d<em>t</em>. Then the value of \(L = \lim_{h \to 0} \frac{1}{h} \int_{1}^{1+2h} e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right) dx\) is</p>
<p>(a) \(e\sin\dfrac{\pi}{3}\)</p>
<p>(b) \(\sqrt{e}\sin\dfrac{\pi}{3}\)</p>
<p>(c) \(e\sin\dfrac{2\pi}{3}\)</p>
<p>(d) \(2e\sin\dfrac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: The limit of an integral over a shrinking interval equals the integrand evaluated at the left endpoint, by the fundamental theorem of calculus. Recognize that ∫₁^(1+2h) f(x)dx ≈ f(1)·2h as h→0, so L = 2f(1).
<p><strong>Step 1:</strong> Recognize that f(x) = e^√x sin(πx/3) is the integrand we need to evaluate.</p><p><strong>Step 2:</strong> Apply the limit definition. As h→0, by the Mean Value Theorem for integrals:</p><p>∫₁^(1+2h) e^√x sin(πx/3) dx ≈ e^√1 sin(π/3) · (2h)</p><p><strong>Step 3:</strong> Therefore:</p><p>L = lim(h→0) [1/h · e^√1 sin(π/3) · 2h]</p><p>L = 2 · e^√1 · sin(π/3)</p><p><strong>Step 4:</strong> Evaluate: sin(π/3) = √3/2 and e^1 = e</p><p>L = 2 · e · (√3/2) = e√3</p><p>∴ Answer: D</p>
Correct Answer: D