The area of the region enclosed by the curves y=x2-4x+4 and y2=16-8x is :
- 83
- 5
- 8
- 43
Answer
(1) 83
Solution
Both parabolas have vertex (2,0); they meet at (2,0) and (0,4) (put y=(x-2)2 in y2=16-8x)
[figure]
For 0 x 2: upper curve 16-8x, lower curve (x-2)2
Area = 02 16-8x dx- 02(x-2)2dx= 163- 83= 83
Answer: (1)