Definite Integration
Properties of Symmetric Functions
Grade 12
Question:
<p>If \(f(a+b+1-x) = f(x)\) for all \(x\), where \(a\) and \(b\) are fixed positive real numbers, then \(\int_{a}^{b} x(f(x) + f(x+1)) dx\) is equal to</p>
<p>(a) \(\int_{a+1}^{b} f(x+1) dx\)</p>
<p>(b) \(\int_{a+1}^{b} f(x) dx\)</p>
<p>(c) \(\int_{a-1}^{b-1} f(x+1) dx\)</p>
<p>(d) \(\int_{a-1}^{b-1} f(x) dx\)</p>
Step-by-Step Solution
Key Concept: Apply the symmetry property $f(a+b+1-x) = f(x)$ combined with the definite integral substitution property to transform the integral.
<p><strong>Solution:</strong> Let $I = \frac{1}{a+b}\int_{a}^{b} x(f(x) + f(x+1)) dx$ ...(i)</p><p>On applying the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we get:</p><p>$I = \frac{1}{a+b}\int_{a}^{b} (a+b-x)(f(a+b-x) + f(a+b-x+1)) dx$</p><p>Using the given condition $f(a+b+1-x) = f(x)$:</p><p>$I = \frac{1}{a+b}\int_{a}^{b} (a+b-x)(f(x+1) + f(x)) dx$ ...(ii)</p><p>Adding (i) and (ii) and simplifying yields $\int_{a-1}^{b-1} f(x+1) dx$</p>
Correct Answer: C