CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The value of 21 f(x) dx, where f(x)=(x+1|+(x|+(x1| is 
  1. 92
  2. 192
  3. 152
  4. 212


Solution

The correct option is B 192
We can redefine f as 
f(x)=   2x,   if 1<x0x+2,   if 0<x1  3x,   if 1<x221f(x) dx=01 f(x) dx+10f(x) dx+21f(x) dx =01(2x) dx+10(x+2) dx+213x dx =(2xx22)01+(x22+2x)10 +3(x22)21=(2+12)+(12+2)+3(212)=192

flag
 Suggest corrections
thumbs-up
 
0 Upvotes



footer-image