Correct Answer - Option 1 : 8/3 square units

**Concept:**

The area of the region bounded by the curve \(\rm f(x)\) , \(\rm x \in [-a, a]\) and the x-axis is given by

A = \(\rm \displaystyle\int_{-a}^a f(x) \;dx \)

**Calculations:**

We know that The area of the region bounded by the curve \(\rm f(x)\) , \(\rm x \in [-a, a]\) and the x-axis is given by

A = \(\rm \displaystyle\int_{-a}^a f(x) \;dx \)

The area of the region bounded by the curve \(\rm f(x)=1 - \dfrac{x^2}{4}, x \in [-2, 2]\) and the x-axis

A = \(\rm \displaystyle\int_{-2}^2 (1 - \dfrac {x^2}{4})dx \)

⇒ A = \(\rm \displaystyle \left[x- \dfrac {x^3}{12}\right]^2_{-2}\)

⇒ A = \(\rm [2 - (-2)]- \left[\frac {8}{12} -\frac {-8}{12} \right]\)

⇒ A = 4 \(-\dfrac 4 3\)

⇒ A = \(\dfrac 8 3\)

Hence, the area of the region bounded by the curve \(\rm f(x)=1 - \dfrac{x^2}{4}, x \in [-2, 2]\) and the x-axis is \(\dfrac 8 3\) square units