Definite Integration
Grade 12
Question:
<p>Let f : [0, 1] <span class="math-tex">\(\to\)</span> R be a continuous function then the maximum value of <span class="math-tex">\(\int_\limits{0}^{1} f(x) \cdot x^{2} d x-\int_\limits{0}^{1} x \cdot(f(x))^{2} d x\)</span> for all such function(s) is :</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{16}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{12}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{8}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{20}\)</span></p>
Step-by-Step Solution
Key Concept: The maximum value is found by completing the square within the integrand to express it as a quadratic in $f(x)$, allowing for pointwise maximization of the integral.
<p><span class="math-tex">$I=\int_{0}^{1}\left(f(x) \cdot x^{2}-x(f(x))^{2}\right) d x$</span><br />
<span class="math-tex">$I=\int_{0}^{1}-x\left(-x f(x)+(f(x))^{2}\right) d x$</span><br />
<span class="math-tex">$I=-\int_{0}^{1} x\left[f(x)^{2}-2 \cdot \frac{x}{2} f(x)+\frac{x^{2}}{4}-\frac{x^{2}}{4}\right] d x$</span><br />
<span class="math-tex">$I=\int_{0}^{1}\left(\frac{x^{3}}{4}-x\left(f(x)-\frac{x}{2}\right)^{2}\right)dx,$</span> which is maximum when <span class="math-tex">$f(x)=\frac{x}{2}$</span><br />
i.e., <span class="math-tex">$I=\int_{0}^{1} \frac{x^{3}}{4} d x=\frac{1}{16}$</span></p>
Correct Answer: A