Definite Integration
Integral + Inequality Chain
Grade 12
Question:
<p>Show \(\displaystyle\int_1^3 e^{x^2-3x}\,dx\le 1\). Find the maximum of \(e^{x^2-3x}\) on \([1,3]\). [JEE Advanced 2016]</p>
1 at x=3/2
e^(-9/4) at x=3/2
1 at x=1 and x=3
e at x=0
Step-by-Step Solution
Key Concept: f(x)=x^2-3x has vertex at x=3/2 (minimum -9/4). At endpoints x=1,3: f=-2. So max of e^f = max(e^(-9/4),e^(-2)) = e^(-2) < 1. Integral \leq 2 \cdot e^(-2) < 1.
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<p>\(f(x)=x^2-3x=(x-3/2)^2-9/4\). Minimum at \(x=3/2\): \(f=-9/4\). At \(x=1,3\): \(f=-2\).</p>
<p>So \(e^{f(x)}\le e^{-2}\) on \([1,3]\) (since \(-9/4 < -2\) and the max occurs at endpoints).</p>
<p>\[\int_1^3 e^{x^2-3x}dx\le e^{-2}\cdot 2=2e^{-2}<1\qquad(\because e^2>2)\]. ✓</p>
<p>The minimum of \(e^{f(x)}\) is \(e^{-9/4}\) at \(x=3/2\). Maximum of integrand is \(e^{-2}\) at endpoints. ✓(B)</p>
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Correct Answer: B