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"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) |
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Answer» `(1)/(2)e^((pi)/(3))log_(e)2` `=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` |
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