Let [.] denote the greatest integer function. Then the value of 03 ( ex+e-x[x]! )dx is :
- e2+e3- 1e2- 1e3
- 12 (e2+e3- 1e2- 1e3 )
- e2+e3- 12e2- 12e3
- 12 (e2+e3 )- 1e2- 1e3
Answer
(2) 12 (e2+e3- 1e2- 1e3 )
Solution
On [0,1): [x]!=0!=1; on [1,2): [x]!=1!=1; on [2,3): [x]!=2!=2
I= 02(ex+e-x) dx+ 12 23(ex+e-x) dx
= [ex-e-x ]02+ 12 [ex-e-x ]23
= (e2-e-2 )+ 12 (e3-e-3-e2+e-2 )
= 12e2+ 12e3- 12e-2- 12e-3
= 12 (e2+e3- 1e2- 1e3 )
Answer: (2)