# What is 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?

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What is 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?
1. 8/3 square units
2. 4/3 square units
3. 2/3 square unit
4. 1/3 square unit
5. None of these

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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