Come trova l'antiderivativo di # e ^ (3x) #?
Risposta:
# int e^(3x) dx = 1/3e^(3x) + C#
Spiegazione:
utilizzando # d/dx e^(ax) = ae^(ax) <=> int ae^(ax) dx = e^(ax) + C'#
# :. int e^(ax) dx = e^(ax)/a + C#
Quindi, # int e^(3x) dx = 1/3e^(3x) + C#
# int e^(3x) dx = 1/3e^(3x) + C#
utilizzando # d/dx e^(ax) = ae^(ax) <=> int ae^(ax) dx = e^(ax) + C'#
# :. int e^(ax) dx = e^(ax)/a + C#
Quindi, # int e^(3x) dx = 1/3e^(3x) + C#