What is #F(x) = int x-xe^(4-2x) dx# if #F(0) = 1 #?

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# integral

#I_1 = intxdx = (x^2)/2 + C_1#

• #I_2# integral

#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#

• F(x) final expression

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#