Question

The area bounded by the curve \(y = 4x^2\), the x-axis, the line x=0 and the line x = 1 is

• A. 2
• B. 2/3
• C. 1/3
• D. 4/3

Hint:

Finding the area under a curve is a direct application of definite integration. Always check if the function is above or below the x-axis in the given interval. If the function dips below the x-axis, you'll need to split the integral and take the absolute value of the negative parts to get the total area.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the area of the region enclosed by the parabola \(y=4x^2\), the x-axis (\(y=0\)), and the vertical lines \(x=0\) and \(x=1\).

Step 2: Key Formula or Approach:
The area under a curve \(y=f(x)\) from \(x=a\) to \(x=b\) is given by the definite integral:
\[ A = \int_{a}^{b} f(x) \,dx \]
We must ensure that \(f(x) \ge 0\) in the interval \([a,b]\).

Step 3: Detailed Explanation:
The function is \(f(x) = 4x^2\).
The boundaries are \(a=0\) and \(b=1\).
In the interval \([0, 1]\), \(x^2\) is always non-negative, so the curve \(y=4x^2\) is above the x-axis.
We set up the definite integral for the area:
\[ A = \int_{0}^{1} 4x^2 \,dx \]
Now, we evaluate the integral:
\[ A = 4 \int_{0}^{1} x^2 \,dx \]
\[ A = 4 \left[ \frac{x^3}{3} \right]_0^1 \]
Evaluate at the limits:
\[ A = 4 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \]
\[ A = 4 \left( \frac{1}{3} - 0 \right) = \frac{4}{3} \]
The area is \(\frac{4}{3}\) square units.

Step 4: Final Answer:
The area of the bounded region is \(\frac{4}{3}\).