`"Given "f(x)=int_(0)^(x)e^(t)(log_(e)sec t- sec^(2)t)dt, g(x)=-2e^(x) tan x,` then the area bounded by the curves `y=f(x) and y=g(x)` between the ordinates `x=0 and x=(pi)/(3),` is (in sq. units)

Options

A. `(1)/(2)e^((pi)/(3))log_(e)2`
B. `e^((pi)/(3))log_(e)2`
C. `(1)/(4)e^((pi)/(3))log_(e)2`
D. `e^((pi)/(3))log_(e)3`

Answer

B

Explanation

`overset(x)underset(0)inte^(t)(log_(e)sec t -sec^(2)t)dt`
`=overset(x)underset(0)inte^(t)[(log _(e) sec t - tan t )+(tan t-sec^(2)t)]dt`
`=[e^(t)(log_(e)sec t-tan t)]_(0)^(x)`
`=e^(x)(log_(e)sec x- tan x)`
`therefore" Required area,"`
`A=overset(pi//3)underset(0)int[e^(x)(log_(e) sec x - tan x)-(-2e^(x)tan x)]dx`
`=overset(pi//3)underset(0)inte^(x)(log_(e)sec x +tan x)]dx`
`=[e^(x)log_(e)sec x]_(0)^(pi//3)`
`=e^(pi//3)log_(e)sec""(pi)/(3)`
`=e^(pi//3)log_(e)2`