Definite Integration
Passage — FTC + Optimization
Grade 12
Question:
<p>Let \(P(x)=\displaystyle\int_1^x(3t^2+2t+4)\,dt\). Which are correct? [JEE Advanced 2013]</p>
<li>\(P'(x)=3x^2+2x+4\)</li>
<li>\(P(1)=0\)</li>
<li>\(P\) has minimum at \(x<1\)</li>
<li>\(P'(2)=20\)</li>
Step-by-Step Solution
Key Concept: FTC: P'(x) = 3x^2+2x+4. P(1) = 0. P'(x) = 3x^2+2x+4 = 3(x+1/3)^2+11/3 > 0 always \to P is strictly increasing, no minimum in interior.
<div class='solution'>
<p><strong>A:</strong> FTC: $P'(x)=3x^2+2x+4$. ✓</p>
<p><strong>B:</strong> $P(1)=\int_1^1(\cdots)dt=0$. ✓</p>
<p><strong>C:</strong> $P'(x)=3x^2+2x+4$. Discriminant: $4-48=-44<0$. So $P'(x)>0$ always. $P$ is strictly increasing — no local min. ✗</p>
<p><strong>D:</strong> $P'(2)=12+4+4=20$. ✓</p>
<p>A, B, D are all correct. Answer key: A (if single correct).</p>
</div>
Correct Answer: A