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

1 Answer
Jan 26, 2016

F(x) = (x^2)/2 + (e^(4-2x))/2(x+1/2) + 1 - (e^4)/4

Explanation:

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