Q.

If the area of the region bounded by the curves y=4–x24 and y=x–42 is equal to α, then 6α equals          [2025]

1 240  
2 220  
3 210  
4 250  

Ans.

(4)

Given, y=4–x24          ... (i)

and y=x–42          ... (ii)

From (i) & (ii), we get

x = –6, 4 and y = –5, 0

Thus, point of intersection of (i) & (ii) are (–6, 5) and (4, 0).

Required Area =∫–64{(4–x24)–(x–42)}dx

⇒α=∫–64{–x24–x2+6}dx

α=–x312–x24+6x|–64

      =–(6412+21612)–(164–364)+6(4+6)

      =1253

∴  6α=250.