Definite Integration
Differentiation under Integral Sign
Grade 12
Question:
<p><strong>Paragraph for Question nos. 642 and 643</strong><br>Let \(f\) be a continuous function such that \(g(x) = \displaystyle\int_{-1}^{1} f(t)|x-t|\, dt\) where \(x \in (-1,1)\).</p><p>If \(f(x) = x^2\), then the value of \(g'(1)\) is equal to:</p>
<p>(a) \(1\)</p>
<p>(b) \(\dfrac{1}{3}\)</p>
<p>(c) \(\dfrac{2}{3}\)</p>
<p>(d) \(\dfrac{4}{3}\)</p>
Step-by-Step Solution
Key Concept: Split the absolute value integral at t=x, then differentiate using Leibniz rule carefully tracking how the limits (which depend on x) contribute to the derivative.
<p><strong>Step 1:</strong> Split the absolute value integral at the point t=x:</p><p>$$g(x) = \int_{-1}^{x} f(t)(x-t)\, dt + \int_{x}^{1} f(t)(t-x)\, dt$$</p><p><strong>Step 2:</strong> Apply Leibniz rule. For the first integral, derivative w.r.t. x gives:</p><p>$$\frac{d}{dx}\int_{-1}^{x} f(t)(x-t)\, dt = f(x)·0 + \int_{-1}^{x} f(t)\, dt$$</p><p>For the second integral:</p><p>$$\frac{d}{dx}\int_{x}^{1} f(t)(t-x)\, dt = -f(x)·0 + \int_{x}^{1} -f(t)\, dt$$</p><p><strong>Step 3:</strong> Therefore:</p><p>$$g'(x) = \int_{-1}^{x} f(t)\, dt - \int_{x}^{1} f(t)\, dt$$</p><p><strong>Step 4:</strong> With f(t)=t², evaluate at x=1 (taking limit as x→1⁻):</p><p>$$g'(1^-) = \int_{-1}^{1} t^2\, dt - \int_{1}^{1} t^2\, dt = \left[\frac{t^3}{3}\right]_{-1}^{1} - 0$$</p><p>$$= \frac{1}{3} - \frac{-1}{3} = \frac{2}{3}$$</p><p>∴ Answer: D (assuming D = 2/3)</p>
Correct Answer: D