Let $f(x) = \int e^x (x-1)(x-2) \, dx$. Then, $f$ decreases in the interval.

Key Concepts Involved

1. Indefinite Integral & Derivative
   If $F(x) = \int g(x)\,dx$, then $F'(x) = g(x)$. The derivative of an antiderivative is just the original integrand.

2. Monotonicity Test
   A function $F(x)$ decreases where its derivative $F'(x)$ is negative.

3. Sign Analysis of Products
   A product $A(x)\,B(x)$ is negative when one factor is positive and the other is negative.

Logical Chain of Thought

1. Find the Derivative
   Because $f(x)$ is an antiderivative, $f'(x) = e^x (x-1)(x-2).$
2. Check where $f'(x)<0$
   • $e^x > 0$ for all real $x$.
   • Therefore, the sign of $f'(x)$ depends only on $(x-1)(x-2)$.
3. Solve $(x-1)(x-2) < 0$
   Roots are at $x=1$ and $x=2$; the parabola opens upwards, hence it is negative between the roots.
   $\boxed{1 < x < 2}$
4. Conclusion
   So $f(x)$ decreases precisely in the open interval $(1,2)$.

What is the question?

We have a function $f(x)$ that is made by integrating (adding up) the expression $e^x (x-1)(x-2)$.

What do we want to know?

We want to know where on the number line $f(x)$ goes downhill (decreases).

How to think about it like a 10-year-old?

1. If you build a LEGO tower one brick at a time, the speed at which the tower grows at any height is like the derivative of the total bricks so far.
2. For our function, that speed is exactly the piece we are adding: $e^x (x-1)(x-2)$.
3. If that speed is negative, the tower actually shrinks—so the total bricks (our $f(x)$) goes down instead of up.
4. $e^x$ is always positive (never negative).
5. So the only way the speed becomes negative is if $(x-1)(x-2)$ is negative.
6. A number line check shows $(x-1)(x-2)$ is negative only between $x=1$ and $x=2$.

👉 Therefore, $f(x)$ is decreasing for all $x$ strictly between 1 and 2.

Step-by-Step Solution

1. Derivative of $f$
   $f'(x) = e^x (x-1)(x-2).$

2. Monotonicity Condition
   $f$ decreases where $f'(x) < 0$. $e^x (x-1)(x-2) < 0.$ Since $e^x > 0\; \forall x$, we simplify to $(x-1)(x-2) < 0.$

3. Solve the Inequality
   • Zeros at $x = 1$ and $x = 2$.
   • Because the quadratic opens upward, it is negative between the roots. $1 < x < 2.$

4. Answer
   $\boxed{f(x)\text{ decreases on }(1,2).}$