The area under the curve in that interval is 8/3 square units.
Let's find the area:
This is actually really simple to do, we just need to integrate the equation:
y = x²
In the range [0, 2]
Because the integral is defined as the area under the curve.
Remember that the integral of the parent quadratic function is:
y = x³/3
Between the range [0, 2] we get the area:
A = [2³/3 - 0²/3]
A = 8/3