Find a value of k > 0 so that the integral from 0 to k of (xe^(-x)) dx = 1. Hint: consider what happens as k approaches infinity.How would I go about solving this? Thanks!

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thilina-g | College Teacher | (Level 1) Educator

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The integral you have to solve is,

`y = int_0^kxe^(-x)dx`

The integral `intxe^(-x)dx` can be found via integration by parts,

`intuv' dt = uv - intvu' dt`

Let,`u = x,` `v' = e^(-x)` , then `v = -e^(-x)` and u' = 1

Therefore,

`intxe^(-x)dx = x(-e^(-x)) - int1 xx (-e^(-x)) dx`

`intxe^(-x)dx = -xe^(-x)-e^(-x)`

`intxe^(-x)dx = -e^(-x)(x+1)`

Therefore,

`y = int_0^kxe^(-x)dx = [-e^(-x)(x+1)]_0^k`

`y = int_0^kxe^(-x)dx = -[e^(-k)(k+1)-1]`

`y = int_0^kxe^(-x)dx = [1-e^(-k)(k+1)]`

This must be equal to 1. Therefore,

`1-e^(-k)(k+1) = 1`

`e^(-k)(k+1) =0`

It is given k>0, therefore `k != -1` .

`(k+1)/e^k = 0`

We can write `e^k` as below according to Taylor series.

`e^k = 1+ k + k^2/2+k^3/6+ ----`

Therefore,


`(k+1)/(1+ k + k^2/2+k^3/6+ ----) = 0`

Divide both numerator and denominator by k.

`(1+1/k)/(1/k+ 1+ k/2+k^2/6+ ----) = 0`

This can only happen if and only if k goes to infinity.

 

Therefore, `k = +oo`

 

`y = int_0^ooxe^(-x)dx = 1`

 

 

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