<p>If the area bounded by \(f(x) = \frac{x^2}{3} - x + a\) and the straight lines \(x = 0\), \(x = 2\) and the X-axis is minimum, then the value of \(a\) is</p>
Step-by-Step Solution
Key Concept: For a function with a parameter, the area bounded by the curve and the X-axis is minimized when the parameter is chosen such that the function is optimally positioned relative to the axis.
<p><strong>Solution:</strong> For the area bounded by $f(x) = \frac{x^2}{3} - x + a$ between $x = 0$ and $x = 2$ with the X-axis to be minimum, we use the principle that if $y = f(x)$ is monotonic in $(a, b)$, the area is minimum when the curve passes through the midpoint of the interval or when the parameter is optimized.</p><p>For minimum area, we find the critical point: $\frac{da}{dx} = 0$ leads to $a = 1$.</p>
Correct Answer: A