Definite Integration
Function Symmetry
Grade 12
Question:
<p>Let <i>f</i> and <i>g</i> be continuous functions on [0, <i>a</i>] such that <i>f</i>(<i>x</i>) = <i>f</i>(<i>a</i> - <i>x</i>) and <i>g</i>(<i>x</i>) + <i>g</i>(<i>a</i> - <i>x</i>) = 4, then <math>∫<sup>a</sup><sub>0</sub> f(x)g(x) dx</math> is equal to</p>
<p>(a) <math>4 \int_0^a f(x) dx</math></p>
<p>(b) <math>\int_0^a f(x) dx</math></p>
<p>(c) <math>2 \int_0^a f(x) dx</math></p>
<p>(d) <math>-3 \int_0^a f(x) dx</math></p>
Step-by-Step Solution
Key Concept: Use the given symmetry conditions: f(x) = f(a-x) and g(x) + g(a-x) = 4 to transform the integral and combine with the original to isolate the answer.
<p>Let I = <math>∫<sup>a</sup><sub>0</sub> f(x)g(x) dx</math>. Using the property that I = <math>∫<sup>a</sup><sub>0</sub> f(a-x)g(a-x) dx</math> and substituting u = a - x, we get I = <math>∫<sup>a</sup><sub>0</sub> f(x)g(a-x) dx</math> (since f(a-x) = f(x)). Adding the original and transformed integrals: 2I = <math>∫<sup>a</sup><sub>0</sub> f(x)[g(x) + g(a-x)] dx = ∫<sup>a</sup><sub>0</sub> f(x) · 4 dx = 4∫<sup>a</sup><sub>0</sub> f(x) dx</math>. Therefore I = <math>2∫<sup>a</sup><sub>0</sub> f(x) dx</math>.</p>
Correct Answer: A