Definite Integration
Leibniz Rule / Limit of Integral
Grade 12

Question:

<p><strong>31.</strong> The value of \(\lim_{h \to 0} \frac{1}{h} \int_{1}^{1+2h} e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right) dx\) equals:</p>
<p>(a) \(\sin\dfrac{\pi}{3}\)</p>
<p>(b) \(4e\sin\dfrac{\pi}{3}\)</p>
<p>(c) \(e\sin\dfrac{\pi}{3}\)</p>
<p>(d) \(2e\sin\dfrac{\pi}{3}\)</p>

Step-by-Step Solution

Key Concept: Recognize this limit as the definition of a derivative: lim(h→0) [F(1+2h) - F(1)]/h = 2F'(1), where F is the antiderivative. By the Fundamental Theorem of Calculus, F'(x) = e^√x sin(πx/3).
<p><strong>Step 1:</strong> Recognize the limit structure as a derivative definition.</p><p>We have: $\lim_{h \to 0} \frac{1}{h} \int_{1}^{1+2h} e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right) dx$</p><p><strong>Step 2:</strong> Use the Fundamental Theorem of Calculus.</p><p>Let $F(x) = \int_1^x e^{\sqrt{t}} \sin\left(\frac{\pi t}{3}\right) dt$</p><p>Then: $\lim_{h \to 0} \frac{F(1+2h) - F(1)}{h}$</p><p><strong>Step 3:</strong> Recognize this as a derivative with a scaling factor.</p><p>$\lim_{h \to 0} \frac{F(1+2h) - F(1)}{h} = \lim_{h \to 0} \frac{F(1+2h) - F(1)}{2h} \cdot 2 = 2F'(1)$</p><p><strong>Step 4:</strong> Apply the Fundamental Theorem.</p><p>$F'(x) = e^{\sqrt{x}} \sin\left(\frac{\pi x}{3}\right)$</p><p>$F'(1) = e^{\sqrt{1}} \sin\left(\frac{\pi}{3}\right) = e \cdot \frac{\sqrt{3}}{2}$</p><p><strong>Step 5:</strong> Calculate the final answer.</p><p>$2F'(1) = 2 \cdot e \cdot \frac{\sqrt{3}}{2} = e\sqrt{3}$</p><p>∴ Answer: D</p>
Correct Answer: D

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