First, we divide the integrate in two parts , that we have called #I_1# and #I_2#:
#I_1 = intxdx#
#I_2 = intxe^(4-2x)dx#
So #F(x)# will be:
#F(x) = I_1 - I_2#
#I_1 = intxdx = (x^2)/2 + C_1#
#I_2 = intxe^(4-2x)dx#
With this integral, we need to use the integration by parts theorem, so we define #u# and #v# as:
#u = x# and its derivative #u' = dx#
#dv = e^(4-2x)dx# and its integrate #v = -1/2e^(4-2x)#
So, #I_2# is:
#I_2 = -x(e^(4-2x))/2 + 1/2inte^(4-2x)dx#
And then:
#I_2 = -x(e^(4-2x))/2 + 1/2(-1/2)e^(4-2x) + C_2#
#I_2 = -(e^(4-2x))/2(x+1/2) + C_2#
We have to subtract both integrals:
#F(x) = I_1 - I_2 = x^2/2 + C_1 + (e^(4-2x))/2(x+1/2) - C_2#
#C_1# and #C_2# are constants so we can group together in one constant that we can define as:
#C = C_1 - C_2#
So, the final expression of #F(x)# is:
#F(x) = x^2/2 + (e^(4-2x))/2(x+1/2) + C#
- Calculation of the constant C
For this, we use the value of #F(x)# in #x=0#: #F(0) = 1#
#F(0) = 0 + e^4/2(0 + 1/2) + C = 1#
#e^4/4 + C = 1#
So, the value of C is:
#C = 1 - e^4/4#
