The area (in square units) of the region {(x,y)∈R|4x2≤y≤8x+12} is
125/3
128/3
124/3
127/3
Step 1. First we draw the graph :
For point of intersection,
⇒4x2=8x+12⇒4x2-8x-12=0⇒4(x2-2x-3)=0⇒x2-2x-3=0⇒x2-3x+x-3=0⇒x(x-3)+1(x-3)=0⇒(x-3)(x+1)=0⇒x=3,-1
Step 2. Calculate the Area bounded by the curves :
A=∫-13(8x+12-4x2)dxA=8x22+12x-4x33-13A=(36+36-36)-(4-12+4/3)=44-43=1283
Hence, the correct option is (B).