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

Jan 26, 2016

$F \left(x\right) = \frac{{x}^{2}}{2} + \frac{{e}^{4 - 2 x}}{2} \left(x + \frac{1}{2}\right) + 1 - \frac{{e}^{4}}{4}$

#### Explanation:

First, we divide the integrate in two parts , that we have called ${I}_{1}$ and ${I}_{2}$:

${I}_{1} = \int x \mathrm{dx}$

${I}_{2} = \int x {e}^{4 - 2 x} \mathrm{dx}$

So $F \left(x\right)$ will be:

$F \left(x\right) = {I}_{1} - {I}_{2}$

• ${I}_{1}$ integral

${I}_{1} = \int x \mathrm{dx} = \frac{{x}^{2}}{2} + {C}_{1}$

• ${I}_{2}$ integral

${I}_{2} = \int x {e}^{4 - 2 x} \mathrm{dx}$

With this integral, we need to use the integration by parts theorem, so we define $u$ and $v$ as:

$u = x$ and its derivative $u ' = \mathrm{dx}$
$\mathrm{dv} = {e}^{4 - 2 x} \mathrm{dx}$ and its integrate $v = - \frac{1}{2} {e}^{4 - 2 x}$

So, ${I}_{2}$ is:

${I}_{2} = - x \frac{{e}^{4 - 2 x}}{2} + \frac{1}{2} \int {e}^{4 - 2 x} \mathrm{dx}$

And then:

${I}_{2} = - x \frac{{e}^{4 - 2 x}}{2} + \frac{1}{2} \left(- \frac{1}{2}\right) {e}^{4 - 2 x} + {C}_{2}$

${I}_{2} = - \frac{{e}^{4 - 2 x}}{2} \left(x + \frac{1}{2}\right) + {C}_{2}$

• F(x) final expression

We have to subtract both integrals:

$F \left(x\right) = {I}_{1} - {I}_{2} = {x}^{2} / 2 + {C}_{1} + \frac{{e}^{4 - 2 x}}{2} \left(x + \frac{1}{2}\right) - {C}_{2}$

${C}_{1}$ 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 $F \left(x\right)$ is:

$F \left(x\right) = {x}^{2} / 2 + \frac{{e}^{4 - 2 x}}{2} \left(x + \frac{1}{2}\right) + C$

• Calculation of the constant C

For this, we use the value of $F \left(x\right)$ in $x = 0$: $F \left(0\right) = 1$

$F \left(0\right) = 0 + {e}^{4} / 2 \left(0 + \frac{1}{2}\right) + C = 1$

${e}^{4} / 4 + C = 1$

So, the value of C is:

$C = 1 - {e}^{4} / 4$