# Use integration by parts to find the given integral. Use symbolic notation and fractions where needed. Integra sign has 1 on top and 0 on bottom....2x(e^(-3x) + e^(-x)) dx

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### 1 Answer

You need to evaluate the definite integral, such that:

`int_0^1 2x(e^(-3x) + e^(-x))dx`

You need to open the brackets such that:

`int_0^1 (2xe^(-3x) + 2xe^(-x))dx`

You need to use the property of linearity of integral, such that:

`int_0^1 (2xe^(-3x) + 2xe^(-x))dx = int_0^1 (2xe^(-3x))dx + int_0^1 (2xe^(-x))dx`

You need to solve the integrals, using integration by parts, such that:

`int_a^b udv = uv|_a^b - int_a^b vdu`

Considering `u = x` and `dv = e^(-3x)dx` yields:

`u = x => du = dx`

`dv = e^(-3x)dx => v = -(e^(-3x))/3`

`int_0^1 (2xe^(-3x))dx = 2int_0^1 (xe^(-3x))dx`

`int_0^1 (xe^(-3x))dx = -x(e^(-3x))/3|_0^1 + (1/3)int_0^1 e^(-3x)dx`

`int_0^1 (xe^(-3x))dx = -x(e^(-3x))/3|_0^1 - (1/9)e^(-3x)|_0^1`

`2int_0^1 (xe^(-3x))dx = 2(-x(e^(-3x))/3|_0^1 - (1/9)e^(-3x)|_0^1)`

`2int_0^1 (xe^(-3x))dx = 2(-1*e^(-3)/3 - e^(-3)/9 + e^0/9)`

`2int_0^1 (xe^(-3x))dx = -2/(3e^3) - 1/(9e^3) + 1/9`

`2int_0^1 (xe^(-3x))dx = -7/(9e^3) + 1/9`

`2int_0^1 (xe^(-3x))dx = (e^3 - 7)/(9e^3)`

Solving the integral `int_0^1 (2xe^(-x))dx` using parts yields:

`u = x => du = dx`

`dv = e^(-x)dx => v = -e^(-x)`

`int_0^1 (2xe^(-x))dx = 2int_0^1 (xe^(-x))dx `

`2int_0^1 (xe^(-x))dx = 2(-xe^(-x)|_0^1 - e^(-x)|_0^1)`

`2int_0^1 (xe^(-x))dx = 2(-1/e - 1/e + e^0) 2int_0^1 (xe^(-x))dx = -4/e + 1 => 2int_0^1 (xe^(-x))dx = (e - 4)/e`

`int_0^1 (2xe^(-3x) + 2xe^(-x))dx = (e^3 - 7)/(9e^3) +(e - 4)/e`

**Hence, evaluating the given definite integral using parts, yields**` int_0^1 (2xe^(-3x) + 2xe^(-x))dx = (e^3 - 7)/(9e^3) +(e - 4)/e.`