The problem asks us to find the function $f(x)$ and the constant $a$ that satisfy the equation $\int_a^x f(t) dt = x^2 - 3x - 2a$.

1. Problem Description

The problem asks us to find the function $f(x)$ and the constant $a$ that satisfy the equation $\int_a^x f(t) dt = x^2 - 3x - 2a$.

2. Solution Steps

To find $f(x)$, we can differentiate both sides of the equation with respect to $x$.

$$\frac{d}{dx} \int_a^x f(t) dt = \frac{d}{dx} (x^2 - 3x - 2a)$$

Using the Fundamental Theorem of Calculus, we have:

$$f(x) = 2x - 3$$

Now, we substitute $x=a$ into the original equation:

$$\int_a^a f(t) dt = a^2 - 3a - 2a$$

Since the integral from $a$ to $a$ is zero, we have:

$$0 = a^2 - 3a - 2a$$

$$0 = a^2 - 5a$$

$$0 = a(a - 5)$$

Therefore, $a=0$ or $a=5$.

If $a=0$, then the equation becomes

$$\int_0^x (2t-3) dt = x^2 - 3x - 2(0)$$

$$[t^2 - 3t]_0^x = x^2 - 3x$$

$$x^2 - 3x - (0^2 - 3(0)) = x^2 - 3x$$

$$x^2 - 3x = x^2 - 3x$$

This solution is valid.

If $a=5$, then the equation becomes

$$\int_5^x (2t-3) dt = x^2 - 3x - 2(5)$$

$$[t^2 - 3t]_5^x = x^2 - 3x - 10$$

$$(x^2 - 3x) - (5^2 - 3(5)) = x^2 - 3x - 10$$

$$x^2 - 3x - (25 - 15) = x^2 - 3x - 10$$

$$x^2 - 3x - 10 = x^2 - 3x - 10$$

This solution is also valid.

Therefore, $f(x) = 2x-3$ and $a=0$ or $a=5$.

3. Final Answer

$f(x) = 2x - 3$

$a = 0$ or $a = 5$