Question

The area (in sq. units) of the region $\left\{\right(𝑥,𝑦)\in 𝑹:{𝑥}^2\le 𝑦\le 3-2𝑥\}$, is

• A. $\frac{31}{3}$
• B. $\frac{32}{3}$
• C. $\frac{29}{3}$
• D. $\frac{34}{3}$

Answer 1

The correct option is B

$\frac{32}{3}$

Explanation for the correct option:

Equations of the curve are:

$$\begin{gathered} y=x^2 \\ y=3-2x \end{gathered}$$

Solving both the equations,

$$\begin{gathered} \Rightarrow x^2=3-2x \\ \Rightarrow x^2-3+2x=0 \\ \Rightarrow x(x+3)-1(x+3)=0 \\ \Rightarrow (x-1)(x+3)=0 \\ \Rightarrow x=1,-3 \end{gathered}$$

Now the area of the required region under the curves is:

$$\begin{gathered} \Rightarrow A={\int }_{-3}^1\left(\right(3-2x)-x^2)dx \\ \Rightarrow A={\int }_{-3}^1(3-2x-x^2)dx \\ \Rightarrow A={\left[\right(3x-x^2-\frac{x^3}{3}\left)\right]}_{-3}^1\Rightarrow A=(3-1-\frac{1}{3}-(-9-9+\frac{27}{3}) \\ \Rightarrow A=((2-\frac{1}{3}+9+9-9) \\ \Rightarrow A=11-\frac{1}{3} \\ \Rightarrow A=\frac{32}{3} \end{gathered}$$

Hence ,option (B) is the correct answer.