Area Under the Curve
Optimization of bounded area
Grade 12
Question:
<p>The value of 'a' (a > 0) for which the area bounded by the curve \(y = \frac{x-1}{6-x^2}\), \(y = 0\), \(x = a\) and \(x = 2a\) has the least value, is</p>
<p>(A) determined by solving optimization conditions</p>
<p>(B) a specific positive value</p>
<p>(C) no value exists</p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Express the area as a function of the parameter a, then differentiate and set equal to zero to find the minimum.
<p>The area $A(a) = \int_a^{2a} \frac{x-1}{6-x^2} dx$. To minimize, compute $\frac{dA}{da} = 0$. Using Leibniz rule and simplifying yields the critical point for the minimum area.</p>
Correct Answer: B