To solve this, we need to evaluate

`int_1^5sqrt(2x-1)\ dx`

Put `u = 2x − 1` then `du = 2 dx`

So `(du)/2=dx`

So we have:

`int_1^5sqrt(2x-1) dx=1/2int_(x=1)^(x=5)u^(1//2)du`

`=1/2xx[2/3u^(3//2)]_(x=1)^(x=5)`

`=1/3[(2x-1)^(3//2)]_1^5`

`=1/3[9^(3//2)-1^(3//2)]`

`=1/3[27-1]`

`=26/3`

So the work done is `8.67` units.

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