Definite Integration
Properties of Definite Integrals — MCQ
Grade None
Question:
<p>Let \(I = \displaystyle\int_0^{\pi} f(x)\,dx\) where \(f(\pi-x)=f(x)\). Which of the following are correct?</p>
I = 2\int_0^(\pi/2) f(x)dx only if f is even
I = 2\int_0^(\pi/2) f(x)dx
\int_0^\pi x \cdot f(x)dx = (\pi/2) \cdot I
\int_0^\pi x \cdot f(x)dx = \pi \cdot \int_0^(\pi/2) f(x)dx
Step-by-Step Solution
Key Concept: If f(\pi-x)=f(x), then \int_0^\pi f = 2\int_0^(\pi/2) f always (no extra condition). King's rule gives \int_0^\pi x \cdot f(x)dx = (\pi/2)I = \pi \cdot \int_0^(\pi/2)f(x)dx.
<div class='solution'>
<p><strong>Claim 1 (A):</strong> False — the condition $f(\pi-x)=f(x)$ alone guarantees $I=2\int_0^{\pi/2}f$. No extra "even" condition is needed.</p>
<p><strong>Claim 2 (B):</strong> True. Sub $x\to\pi-x$: $I=\int_0^\pi f(\pi-x)dx=\int_0^\pi f(x)dx=I$. Split at $\pi/2$:</p>
<p>$$I=\int_0^{\pi/2}f(x)dx+\int_{\pi/2}^\pi f(x)dx\stackrel{x\to\pi-x}{=}2\int_0^{\pi/2}f(x)dx$$</p>
<p><strong>Claim 3 (C):</strong> True. Let $J=\int_0^\pi xf(x)dx$. King: $J=\int_0^\pi(\pi-x)f(x)dx$. Add: $2J=\pi I\Rightarrow J=\frac{\pi}{2}I$.</p>
<p><strong>Claim 4 (D):</strong> True. $J=\frac{\pi}{2}I=\frac{\pi}{2}\cdot2\int_0^{\pi/2}f=\pi\int_0^{\pi/2}f(x)dx$.</p>
</div>
Correct Answer: ['B', 'C', 'D']