Let ƒ(x) = ^x∫₀  e^t(t-1)(t-2) dt. Then ƒ(x) decreases in the interval

The correct option is x ∈ (1,2).

Step-by-step Explanation:

We are given the function defined by the integral:

$f\left(x\right)={\int }_0^x{e}^t(t-1)(t-2)dt$

To determine the interval in which the function $f\left(x\right)$ is decreasing, we need to find the interval where its first derivative is negative, i.e., $f\text{'}\left(x\right)<0$.

Using the Leibniz Rule for differentiating under the integral sign, we differentiate $f\left(x\right)$ with respect to $x$:

$f\text{'}\left(x\right)=\frac{d}{dx}\left[{\int }_0^x{e}^t\right(t-1\left)\right(t-2\left)dt\right]$

$f\text{'}\left(x\right)=e^x(x-1)(x-2)$

For the function to be decreasing, we set the derivative to be less than zero:

$e^x(x-1)(x-2)<0$

Since the exponential function $e^x$ is strictly positive for all real numbers $x$, we can divide both sides of the inequality by $e^x$ without changing the direction of the inequality:

$(x-1)(x-2)<0$

To solve this inequality, we analyze the signs of the factors in the intervals created by the roots $x=1$ and $x=2$:
- If $x<1$, both $(x-1)$ and $(x-2)$ are negative, making their product positive.
- If $x>2$, both $(x-1)$ and $(x-2)$ are positive, making their product positive.
- If $1<x<2$, the factor $(x-1)$ is positive and $(x-2)$ is negative, making their product negative.

Therefore, the inequality holds true when:

$x\in (1,2)$

Thus, $f\left(x\right)$ decreases in the interval (1, 2).