The user wants me to analyze the function $f\left( x \right) = \int_0^x {{e^{{t^2}}}} \left( {t - 2} \right)\left( {t - 3} \right)dt$ for $x \in \left( {0,\infty } \right)$ and determine its properties regarding local extrema and intervals of monotonicity.

Step 1: Find the first derivative of f(x)

To analyze the function's behavior, we first need to find its derivative, $f'(x)$. We can use the First Fundamental Theorem of Calculus (also known as Leibniz's rule for differentiation of an integral). If $F(x) = \int_a^x g(t) dt$, then $F'(x) = g(x)$.

In our case, the integrand is $g(t) = e^{t^2}(t - 2)(t - 3)$. So, the derivative of $f(x)$ is: $f'(x) = \frac{d}{{dx}}\left( {\int_0^x {{e^{{t^2}}}} \left( {t - 2} \right)\left( {t - 3} \right)dt} \right) = {e^{{x^2}}}\left( {x - 2} \right)\left( {x - 3} \right)$

Step 2: Find the critical points

Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. The function $f'(x)$ is defined for all real numbers, so we only need to find where $f'(x) = 0$.

${e^{{x^2}}}\left( {x - 2} \right)\left( {x - 3} \right) = 0$

Since $e^{x^2}$ is always positive for any real $x$, the equation is satisfied only when: $\left( {x - 2} \right)\left( {x - 3} \right) = 0$

This gives us the critical points $x = 2$ and $x = 3$. Both of these critical points lie within the given domain of $f(x)$, which is $(0, \infty)$.

Step 3: Analyze the sign of f'(x) to determine intervals of increase/decrease

The sign of $f'(x)$ depends on the sign of the quadratic factor $(x - 2)(x - 3)$, as $e^{x^2}$ is always positive. We analyze the sign of $f'(x)$ in the intervals defined by the critical points within the domain $(0, \infty)$: $(0, 2)$, $(2, 3)$, and $(3, \infty)$.

• Interval (0, 2): For any $x$ in this interval, say $x=1$, $(x-2)$ is negative and $(x-3)$ is negative. Their product is positive. Thus, $f'(x) > 0$, which means $f(x)$ is increasing on $(0, 2)$.
• Interval (2, 3): For any $x$ in this interval, say $x=2.5$, $(x-2)$ is positive and $(x-3)$ is negative. Their product is negative. Thus, $f'(x) < 0$, which means $f(x)$ is decreasing on $(2, 3)$.
• Interval (3, ∞): For any $x$ in this interval, say $x=4$, $(x-2)$ is positive and $(x-3)$ is positive. Their product is positive. Thus, $f'(x) > 0$, which means $f(x)$ is increasing on $(3, \infty)$.

Step 4: Evaluate each option based on the analysis

• Option A: $f$ has a local maximum at $x=2$ At $x=2$, the derivative $f'(x)$ changes its sign from positive to negative. According to the First Derivative Test, this indicates a local maximum. Thus, option A is correct.

• Option B: $f$ is decreasing on $(2, 3)$ Our analysis in Step 3 showed that for $x \in (2, 3)$, $f'(x) < 0$. This means the function $f(x)$ is indeed decreasing on the interval $(2, 3)$. Thus, option B is correct.

• Option C: there exists some $c \in \left( {0,\infty } \right),$ such that $f'(c)=0$ We found two critical points, $x=2$ and $x=3$, by solving $f'(x)=0$. Since both $2$ and $3$ are in the interval $(0, \infty)$, such a value $c$ exists (in fact, there are two such values). Thus, option C is correct.

• Option D: $f$ has a local minimum at $x=3$ At $x=3$, the derivative $f'(x)$ changes its sign from negative to positive. According to the First Derivative Test, this indicates a local minimum. Thus, option D is correct.

Conclusion

All four statements A, B, C, and D are correct descriptions of the function $f(x)$.