Use integration by parts to find given integral. Use symbolic notation and fractions where needed.  `int_0^1 2x(e^(-3x) + e^(-x))dx`

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lemjay eNotes educator| Certified Educator

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

The formula of integration by parts is `int udv = u*v- int vdu` .

So let,

`u=2x`                   and               `dv=(e^(-3x) + e^(-x))dx`

Then, determine du and v.

`du=2dx`             and                   `v=int (e^(-3x) + e^(-x))dx`

                                                     `v= -e^(-3x)/3 - e^(-x)`

Plug-in u, v and du to the formula of integration by parts.


`= 2x*(-e^(-3x)/3-e^(-x)) - int (-e^(-3x)/3-e^(-x))*2dx`

`=2x(-e^(-3x)/3-e^(-x) ) +2int(e^(-3x)/3+e^(-x))dx`


Apply distributive property to remove the parenthesis.

`=-(2xe^(-3x))/3 - 2xe^(-x) -(2e^(-3x))/9-2e^(-x)`

Then, plug-in the limits of the integral.

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

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

`=(-(2*1*e^(-3*1))/3 - 2*1*e^(-1) -(2e^(-3*1))/9-2e^(-1)) `

    `-(-(2*0*e^(-3*0))/3 - 2*0*e^(0) -(2e^(-3*0))/9-2e^(0))`

`= (-(2e^(-3))/3 -2e^(-1) -(2e^(-3))/9-2e^(-1) ) -(0-0-(2e^0)/9-2e^0)`

`=(-(2e^(-3))/3 - 2e^(-1) -(2e^(-3))/9-2e^(-1) )- (-2/9-2)`

`=-(2e^(-3))/3-4e^(-1)-(2e^(-3))/9 +2/9+2`

Re-write the fractions using the LCD as denominator.


`=-(8e^(-3))/9 -(36e^(-1))/9+20/9`

`=20/9-(8e^(-3)+36e^(-1))/9 `

Hence, `int_0^1 2x(e^(-3x) + e^(-x))dx=20/9-(8e^(-3)+36e^(-1))/9` .