Here is the area under consideration:

In this case, `y = f(x) = x^3`, `a = 0`, `b = 2`.

We find the shaded area first:

`A=int_0^2 x^3 dx = [(x^4)/(4)]_0^2=16/4=4`

Next, using the formula for the *x*-coordinate of the centroid we have:

`barx=1/A int_a^b xf(x) dx`

`=1/4 int_0^2 x(x^3) dx`

`=1/4 int_0^2 (x^4) dx`

`=1/4[(x^5)/(5)]_0^2`

`=32/20`

`=1.6`

Now, for the *y* coordinate, we need to find:

`x_2 = 2` (this is fixed in this problem)

`x_1 = y^(1//3)` (this is variable in this problem)

`c = 0`, `d = 8`.`bary=1/A int_c^d y(x_2-x_1) dy`

`=1/4 int_0^8 y(2-y^[1/3]) dy`

`=1/4 int_0^8(2y - y^[4/3]) dy`

`=1/4[y^2-(3y^[7/3])/(7)]_0^8`

`=1/4 [64-(3 times 128)/(7)]`

`=2.29`

So the centroid for the shaded area is at (1.6, 2.29).

**Alternate Method for the y-coordinate**

Using the "Method 2" formula given, we could also obtain the *y*-coordinate of the centroid as follows:

`bary=1/A int_a^b ([f(x)]^2)/(2) dx`

`=1/4 int_0^2 ([x^3]^2)/(2) dx`

`=1/4 int_0^2 (x^6)/(2) dx`

`=1/56 [x^7]_0^2`

`=2.29`

In this example, Method 2 is easier than Method 1, but it may not always be the case.

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