The function $$f\left( x \right) = \int_{ - 1}^x {t\left( {{e^t} - 1} \right){{\left( {t - 2} \right)}^3}{{\left( {t - 3} \right)}^5}} dt$$         has a local minimum at $$x$$ which is equal to :

Solution :
$$\eqalign{ & f'\left( x \right) = x\left( {{e^x} - 1} \right){\left( {x - 2} \right)^3}.{\left( {x - 3} \right)^5} \cr & f'\left( {0 - \in } \right) = \left( {0 - \in } \right)\left( {{e^{0 - \in }} - 1} \right){\left( {0 - \in - 2} \right)^3}{\left( {0 - \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {0 + \in } \right) = \left( {0 + \in } \right)\left( {{e^{0 + \in }} - 1} \right){\left( {0 + \in - 2} \right)^3}{\left( {0 + \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {1 - \in } \right) = \left( {1 - \in } \right)\left( {{e^{1 - \in }} - 1} \right){\left( {1 - \in - 2} \right)^3}{\left( {1 - \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {1 + \in } \right) = \left( {1 + \in } \right)\left( {{e^{1 + \in }} - 1} \right){\left( {1 + \in - 2} \right)^3}{\left( {1 + \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {2 - \in } \right) = \left( {2 - \in } \right)\left( {{e^{2 - \in }} - 1} \right){\left( {2 - \in - 2} \right)^3}{\left( {2 - \in - 3} \right)^5} = \left( { + {\text{ve}}} \right) \cr & f'\left( {2 + \in } \right) = \left( {2 + \in } \right)\left( {{e^{2 + \in }} - 1} \right){\left( {2 + \in - 2} \right)^3}{\left( {2 + \in - 3} \right)^5} = \left( { - {\text{ve}}} \right) \cr & {\text{so, a local maximum at }}x = 2 \cr & f'\left( {3 - \in } \right) = \left( {3 - \in } \right)\left( {{e^{3 - \in }} - 1} \right){\left( {3 - \in - 2} \right)^3}{\left( {3 - \in - 3} \right)^5} = \left( { - {\text{ve}}} \right) \cr & f'\left( {3 + \in } \right) = \left( {3 + \in } \right)\left( {{e^{3 + \in }} - 1} \right){\left( {3 + \in - 2} \right)^3}{\left( {3 + \in - 3} \right)^5} = \left( { + {\text{ve}}} \right); \cr & {\text{so, a local minimum at }}x = 3 \cr} $$