| 1. |
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 units2. 4/3 square units3. 2/3 square unit4. 1/3 square unit |
|
Answer» 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
|
|