Question

The area (in square units) of the region ${\left\{\right(x,y)\in R|4x^2\le y\le 8x+12\}}^{}$ is

• A. $125/3$
• B. $128/3$
• C. $124/3$
• D. $127/3$

Answer 1

The correct option is B

$128/3$

Step 1. First we draw the graph :

For point of intersection,

$$\begin{gathered} \Rightarrow 4x^2=8x+12 \\ \Rightarrow 4x^2-8x-12=0 \\ \Rightarrow 4(x^2-2x-3)=0 \\ \Rightarrow x^2-2x-3=0 \\ \Rightarrow x^2-3x+x-3=0 \\ \Rightarrow x(x-3)+1(x-3)=0 \\ \Rightarrow (x-3)(x+1)=0 \\ \Rightarrow x=3,-1 \end{gathered}$$

Step 2. Calculate the Area bounded by the curves :

$$\begin{gathered} A={\int }_{-1}^3(8x+12-4x^2)dx \\ A={\left[\frac{8x^2}{2}+12x-\frac{4x^3}{3}\right]}_{-1}^3 \\ A=(36+36-36)-(4-12+4/3) \\ =44-\frac{4}{3} \\ =\frac{128}{3} \end{gathered}$$

Hence, the correct option is (B).