Given f(x)=x^5+2x-1, what is `int_0^1` f(x)dx?

You need to evaluate the definite integral, such that:

`int_0^1 f(x)dx = int_0^1 (x^5 + 2x - 1)dx`

Using the property of linearity of integral, you need to split the given integral in three simpler integrals, such that:

`int_0^1 f(x)dx = int_0^1 x^5 dx + int_0^1 2x dx - int_0^1 dx`

`int_0^1 f(x)dx = (x^6/6 + 2x^2/2 - x)|_0^1`

Reducing duplicate factors yields:

`int_0^1 f(x)dx = (x^6/6 + x^2 - x)|_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 f(x)dx = (1^6/6 + 1^2 - 1 - 0^6/6 - 0^2 + 0)`

`int_0^1 f(x)dx = 1/6`

Hence, evaluating the given definite integral, yields `int_0^1 f(x)dx = 1/6.`