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).

To find the interval in which the function
$f\left(x\right)={\int }_0^x{e}^t(t-1)(t-2)dt$
decreases, we need to determine 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 get:
$f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[{\int }_0^x{e}^t(t-1)(t-2)dt\right]$
$f^{\text{'}}\left(x\right)=e^x(x-1)(x-2)$

For the function to be decreasing, we set
$f^{\text{'}}\left(x\right)<0$:
$e^x(x-1)(x-2)<0$

Since the exponential term
$e^x$
is always positive for all real values of x, the sign of the derivative is entirely determined by the quadratic factor:
$(x-1)(x-2)<0$

Solving this inequality, the product is negative when x lies strictly between the roots of the equation:
$1<x<2$

Thus, the function decreases in the interval x ∈ (1, 2).