If `f(x)=int_0^1e^|t-x|dt` where `(0<=x<=1)`, then maximum value of `f(x)` is

If f(x)=int_(0)^(1)e^(|t-x|)dt where (0<=x<=1) then maximum value of f(x) is

If f(x)=int_(0)^(x)|t-1|dt , where 0lexle2 , then

Let f(x)=int_(0)^(1)|x-t|dt, then

Find the range of the function f (x) = int _(0) ^(x) |t-1| dt, where 0 le x le 2

If f(x)=int_(0)^(x)e^(-t)f(x-t)dt then the value of f(3) is

A Function f(x) satisfies the relation f(x)=e^(x)+int_(0)^(1)e^(x)f(t)dt* Then (a)f(0) 0

For the function f(x)=int_(0)^(x)(sin t)/(t)dt where x>0

A continuous function f(x) satisfies the relation f(x)=e^(x)+int_(0)^(1)e^(x)f(t)dt then f(1)=

Let f ( x) = int _ 0 ^ x g (t) dt , where g is a non - zero even function . If f(x + 5 ) = g (x) , then int _ 0 ^ x f (t ) dt equals :

f(x)=x^(3)+1-2x int_(0)^(x)g(t)dt and g(x)=x-int_(0)^(1)f(t)dt. Then yhe value of int_(0)^(1)f(t)dt lies in the interval