Let $a$ be positive numbers.For $|x|\leq a,$ find the maximum and minimum value of $\int_{x-a}^{x+a}\sqrt{4a^{2}-t^{2}}\ dt.$
1. Define the function \( f(x) = \int_{x-a}^{x+a} \sqrt{4a^2 - t^2} \, dt \).

2. To find the critical points, we need to compute the derivative \( f'(x) \). Using the Fundamental Theorem of Calculus and the Chain Rule, we get:
   \[
   f'(x) = \sqrt{4a^2 - (x+a)^2} - \sqrt{4a^2 - (x-a)^2}
   \]

3. Analyze the sign of \( f'(x) \):
   - For \( x \leq 0 \):
     \[
     f'(x) = \sqrt{4a^2 - (x+a)^2} - \sqrt{4a^2 - (x-a)^2} \geq 0
     \]
     This implies that \( f(x) \) is non-decreasing for \( x \leq 0 \).

   - For \( x \geq 0 \):
     \[
     f'(x) = \sqrt{4a^2 - (x+a)^2} - \sqrt{4a^2 - (x-a)^2} \leq 0
     \]
     This implies that \( f(x) \) is non-increasing for \( x \geq 0 \).

4. Since \( f(x) \) is non-decreasing for \( x \leq 0 \) and non-increasing for \( x \geq 0 \), the maximum value of \( f(x) \) occurs at \( x = 0 \).

5. Compute \( f(0) \):
   \[
   f(0) = \int_{-a}^{a} \sqrt{4a^2 - t^2} \, dt
   \]
   This integral represents the area of a semicircle with radius \( 2a \). The area of a semicircle is given by:
   \[
   \int_{-a}^{a} \sqrt{4a^2 - t^2} \, dt = \frac{1}{2} \pi (2a)^2 = 2\pi a^2
   \]

6. The minimum value of \( f(x) \) occurs at the endpoints \( x = \pm a \).

7. Compute \( f(a) \) and \( f(-a) \):
   \[
   f(a) = \int_{0}^{2a} \sqrt{4a^2 - t^2} \, dt
   \]
   This integral also represents the area of a semicircle with radius \( 2a \), but only over the interval from \( 0 \) to \( 2a \). The area of this quarter circle is:
   \[
   \int_{0}^{2a} \sqrt{4a^2 - t^2} \, dt = \frac{1}{4} \pi (2a)^2 = \pi a^2
   \]

8. Therefore, the maximum value of \( f(x) \) is \( 2\pi a^2 \) and the minimum value of \( f(x) \) is \( \pi a^2 \).

The final answer is \(\boxed{2\pi a^2}\) for the maximum value and \(\boxed{\pi a^2}\) for the minimum value.