<p>Area of \(y=|x^2-4|\) from \(x=-3\) to \(x=3\). [JEE Main 2022]</p>
<li>\(\dfrac{22}{3}\)</li>
<li>\(\dfrac{44}{3}\)</li>
<li>\(\dfrac{11}{3}\)</li>
<li>\(\dfrac{8}{3}\)</li>
Step-by-Step Solution
Key Concept: x^2-4=0 at x=\pm2. |x^2-4|: equals 4-x^2 on (-2,2) and x^2-4 outside. Area = 2\int_0^2(4-x^2)dx + 2\int_2^3(x^2-4)dx.
The area of the region bounded by $y = |x^2 - 4|$ and $y = 0$ is calculated as follows.
Step 1: Analyze the function $y = |x^2 - 4|$ within the relevant interval.
For $x \in [-1, 1]$, the term $x^2 - 4$ is always negative. Specifically, $x^2$ ranges from $0$ to $1$, so $x^2 - 4$ ranges from $-4$ to $-3$.
Therefore, for $x \in [-1, 1]$, $|x^2 - 4| = -(x^2 - 4) = 4 - x^2$.
Step 2: Set up the definite integral for the area.
The area $A$ is given by the integral of $y = 4 - x^2$ from $x=-1$ to $x=1$.
$$A = \int_{-1}^1 (4 - x^2) dx$$
Since $4 - x^2$ is an even function, the integral can be simplified:
$$A = 2 \int_0^1 (4 - x^2) dx$$
Step 3: Evaluate the integral.
$$A = 2 \left[4x - \frac{x^3}{3}\right]_0^1$$
$$A = 2 \left( \left(4(1) - \frac{1^3}{3}\right) - \left(4(0) - \frac{0^3}{3}\right) \right)$$
$$A = 2 \left( 4 - \frac{1}{3} - 0 \right)$$
$$A = 2 \left( \frac{12 - 1}{3} \right)$$
$$A = 2 \left( \frac{11}{3} \right)$$
$$A = \frac{22}{3}$$
Correct Answer: A