A) \[\log 4\]sq. unit
B) \[(\log 4+1)\]sq. unit
C) \[(\log 4-1)\]sq. unit
D) None of these
Correct Answer: C
Solution :
Given curve \[y=\log x\] and \[x=1\], \[x=2\]. Hence required area = \[\int_{1}^{2}{\,\log \,x\,dx}\] = \[(x\,\log x-x)_{1}^{2}\] = \[2\log 2-1=(\log 4-1)\]sq. unit.You need to login to perform this action.
You will be redirected in
3 sec