The second fundamental theorem of calculus states that for some function #f(x)# that is continuous over the interval #[a, b]# (where #a# is a constant), there exists a function #F(x) = int_(a)^(x) f(t) d t#.
This means that #F(x)# is an antiderivative of #f(x)#, or #F'(x) = f(x),# for all #x# in #b#.
#Rightarrow frac(d)(dx) (int_(a)^(x) f(t) d t) = f(x)#
Let's substitute our function into this form:
#Rightarrow frac(d)(dx) (int_(- 3)^(x) (t^(2) + 3 t + 2) d t) = x^(2) +3 x + 2#
#therefore F'(x) = x^(2) + 3 x + 2#
Now that we have #F'(x)#, we can evaluate its antiderivative, which will give us #F(x)#.
First, let's evaluate the indefinite integral of #F'(x)#:
#Rightarrow int(x^(2) + 3 x + 2) d x = int(x^(2)) d x + int(3 x) d x + int(2) d x#
#Rightarrow int(x^(2) + 3 x + 2) d x = frac(x^(2 + 1))(2 + 1) + frac(3 x^(1 + 1))(1 + 1) + 2 x + C#
#therefore int(x^(2) + 3 x + 2) d x = frac(x^(3))(3) + frac(3 x^(2))(2) + 2 x + C#
Then, let's calculate the definite integral in the interval #[- 3, x]#:
#Rightarrow int_(- 3)^(x) (x^(2) + 3 x + 2) d x = F(x) - F(- 3)#
Now, #F( -3) = int_(- 3)^(- 3) (x^(2) + x + 2) d x = 0#, because the integration is being done over an interval of length #0#:
#Rightarrow int_(- 3)^(x) (x^(2) + 3 x + 2) d x = (frac(x^(3))(3) + frac(3 x^(2))(2) + 2 x) - (0)#
#Rightarrow int_(- 3)^(x) (x^(2) + 3 x + 2) d x = frac(x^(3))(3) + frac(3 x^(2))(2) + 2 x#
#Rightarrow int_(- 3)^(x) (x^(2) + 3 x + 2) d x = frac(x^(3))(3) + frac(3)(2) x^(2) + 2 x#
#therefore F(x) = frac(x^(3))(3) + frac(3)(2) x^(2) + 2 x#
