Area bounded by the curves y = x ,y ` = tanx and x = pi/4` is

Options

A. `pi/4` sq unit
B. `(log sqrt2 = (pi^(2))/32)` sq unit
C. `(log2 - (pi^(2))/16)` sq unit
D. none of these

Answer

B

Explanation

We have, y= x,y = tan x and ` x = pi/4`
Since ` tan x gt x in [ 0, pi/4]`
Area bounded by curve y =x
` y = tan x and x = pi/4 ` is
`int_(0)^(pi//4) (tanx -x)dx`
` [ log secx- x^(2)/2]_(0)^(pi//4)`
` = log sec pi/4 - pi^(2)/32`
` ( log sqrt2 - pi^(2)/32)`