let f(x)=∫_0 ^(+∞) ((arctan(xt))/(1+t^2 ))dt with x≥0
1) calculate f^′ (x) then a simple form of f(x)
2) calculate ∫_0 ^(+∞) ((arctant)/(1+t^2 ))dt
3) calculate ∫_0 ^(+∞) ((arctan(2t))/(1+t^2 ))dt
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(i) given the function f(t)=e^t and g(t)=lnt
show that f○g(t)=g○f(t)
(ii)if f(t)=at , g(t)=bt^2 +3
(fog)(2)=35 and (fog)(3)=75
find the value of a and b
It is given that the first term of
a GP is the last term of an AP.
the second term of the AP is the
third term of the GP..detemine
the Geometric mean of the GP is
the fourth term of the GP is 16.
A man 2m 50cm tall stands a
distance of 3m in front of a large
vertical plane mirror.
i)what is the shortest length of the
mirror that will enable the man see
himself fully?
ii)what is the answer of the above
if the man were 5m away?
let x>0 and F(x)= ∫_0 ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt
1) find a simple form of F(x)
2)find the value of ∫_0 ^∞ ((arctan(2t^2 ))/(1+t^2 ))dt
3)find the value of ∫_0 ^∞ ((arctan(3t^2 ))/(1+t^2 ))dt.