Step 1: Understanding the Concept:
We evaluate the integral first and then analyze the resulting equation. A solution exists if the range of the left-hand side matches the possible values of the right-hand side.
Step 2: Key Formula or Approach:
1. Evaluate the definite integral.
2. Solve the equation for \( \sin(a/x) \).
3. Use the property that \( |\sin \theta| \le 1 \).
Step 3: Detailed Explanation:
Evaluate the integral:
\[ \int_{0}^{x} (t^2 - 8t + 13) \, dt = [ \frac{t^3}{3} - 4t^2 + 13t ]_{0}^{x} = \frac{x^3}{3} - 4x^2 + 13x \]
The equation is:
\[ \frac{x^3}{3} - 4x^2 + 13x = x \sin \frac{a}{x} \]
Divide by \( x \) (assuming \( x \neq 0 \)):
\[ \frac{x^2}{3} - 4x + 13 = \sin \frac{a}{x} \]
For a solution to exist, the quadratic \( Q(x) = \frac{x^2}{3} - 4x + 13 \) must be in the range of the sine function, which is \( [-1, 1] \).
Let's find the minimum value of the quadratic \( Q(x) \):
The vertex is at \( x = \frac{-b}{2a} = \frac{4}{2/3} = 6 \).
\( Q(6) = \frac{36}{3} - 4(6) + 13 = 12 - 24 + 13 = 1 \).
Since the quadratic opens upwards and its minimum value is 1, the only way for the equation to have a solution is if the quadratic equals exactly 1.
So, \( \sin \frac{a}{x} = 1 \) at \( x = 6 \).
\[ \frac{a}{6} = (2n + \frac{1}{2})\pi \implies a = 6(2n + \frac{1}{2})\pi \]
For the least positive value of \( a \), we set \( n = 0 \) or \( n=1 \) appropriately.
Wait, let's check \( \sin \theta = 1 \implies \theta = \pi/2, 5\pi/2, \dots \).
If \( a/6 = \pi/2 \), \( a = 3\pi \).
If we used \( x \) values that make the quadratic slightly larger than 1, there would be no real \( a/x \). Since 1 is the only overlapping value between the range of \( Q(x) \) and \( \sin(\theta) \), \( x=6 \) is required.
Step 4: Final Answer:
The least positive value of \( a \) is \( 3\pi \).