Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>We have \(\frac{d}{dx}F(x) = \left(\frac{e^{\sin x}}{x}\right),\ x > 0\). If \(I = \int_1^4 \frac{3}{x} e^{\sin x^3} dx\), then \(I\) equals:</p>
<p>\(F(64) - F(1)\)</p>
<p>\(F(64) + F(1)\)</p>
<p>\(3[F(64) - F(1)]\)</p>
<p>\(K = 64\)</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand matches a derivative form through substitution u = x³, which transforms the given derivative condition into the exact form of I. This reveals I = F(4) - F(1) without computing F explicitly.
<p><strong>Step 1:</strong> Given that F'(x) = e^(sin x)/x for x > 0, we need to express I in terms of F.</p><p><strong>Step 2:</strong> In integral I = ∫₁⁴ (3/x)e^(sin x³) dx, use substitution u = x³, so du = 3x² dx.</p><p><strong>Step 3:</strong> Rewrite: (3/x)e^(sin x³) dx = (3x²/x³)e^(sin x³) dx = e^(sin x³)/u · du where u = x³.</p><p><strong>Step 4:</strong> When x = 1: u = 1; when x = 4: u = 64. This gives I = ∫₁⁶⁴ e^(sin u)/u du = F(64) - F(1).</p><p><strong>Correction:</strong> Substitution u = x³: du = 3x² dx, so (3/x)e^(sin x³) dx requires rewriting. More directly: letting u = x³ gives I = ∫₁⁶⁴ (e^(sin u)/u) du = F(64) - F(1).</p><p><strong>Step 5:</strong> By Fundamental Theorem of Calculus: I = F(4) - F(1).</p><p>∴ Answer: A</p>
Correct Answer: A