Step 1: Understanding the Concept:
The area under a curve $y = f(x)$ from $x=a$ to $x=b$ (and above the x-axis) is given by the definite integral of the function over that interval.
Step 2: Key Formula or Approach:
The formula for the area is:
\[ A = \int_a^b f(x) \,dx \]
In this problem, $f(x) = 4x^2$, $a=0$, and $b=1$. The curve $y=4x^2$ is always non-negative, so it lies above the x-axis in the given interval.
Step 3: Detailed Explanation:
We need to calculate the area A bounded by $y = 4x^2$, the x-axis ($y=0$), $x=0$, and $x=1$.
Set up the definite integral:
\[ A = \int_0^1 4x^2 \,dx \]
Use the power rule for integration $\int x^n \,dx = \frac{x^{n+1}}{n+1}$:
\[ A = 4 \int_0^1 x^2 \,dx = 4 \left[ \frac{x^3}{3} \right]_0^1 \]
Now, evaluate the integral at the upper and lower limits:
\[ A = 4 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \]
\[ A = 4 \left( \frac{1}{3} - 0 \right) \]
\[ A = \frac{4}{3} \]
Step 4: Final Answer:
The area bounded by the curve and the lines is $\frac{4}{3}$ square units. Therefore, option (D) is correct.