Please help me with this challenging problem: Find the area enclosed between the curve y = x², x-axis, and x = 4. (A sketch of the curve is required.)

To find the area enclosed between the curve y = x², x-axis, and x = 4, we need to calculate the definite integral of the function y = x² from x = 0 to x = 4.

First, let's sketch the curve y = x²:

```
| *
| * *
| * *
| * *
| * *
| * *
| * *
|* *
-----------------
0 1 2 3 4
```

Now, we need to calculate the definite integral of y = x² from x = 0 to x = 4:

∫[0,4] x² dx = [1/3 x³] [0,4] = 1/3 * 4³ - 1/3 * 0³ = 64/3

Therefore, the area enclosed between the curve y = x², x-axis, and x = 4 is 64/3 square units.

[Sketch of the Curve]

[Image of the curve y = x^2, the x-axis, and the vertical line x = 4]

[Solution]

To find the area enclosed between the curve and the x-axis, we need to integrate the function y = x^2 over the interval [0, 4].

```
Area = ∫[0,4] x^2 dx
```

Evaluating the integral, we get:

```
Area = [x^3/3] from 0 to 4
Area = (4^3/3) - (0^3/3)
Area = 64/3
Area = 21.33 (rounded to two decimal places)
```

Therefore, the area enclosed between the curve y = x², the x-axis, and the vertical line x = 4 is approximately 21.33 square units.