Sketch first:

Graph of `y=x^3`, showing the portion bounded by `x = 0` and `y=3`.

We will use:

`text[Area]=int_c^df(y) dy`

and use **horizontal elements**. (In this example we could have added horizontally as well, but will do it vertically to illustrate the method.)

In this case, `c = 0` and `d = 3`.

We need to express `x` in terms of `y`:

`y = x^3` so `x = y^(1//3)`

So

`text[Area]=int_c^df(y) dy`

`=int_0^3 (y^(1//3)) dy`

`=[3/4 y^(4//3)]_0^3`

`=3/4 [(3)^(4//3)-(0)^(4//3)]`

`=3.245\ text[sq units]`

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