If `f(x)={ int_0^x |1-t|dt ; x>1 , x-1/2 ; x<=1` then

let f(x)={( int_0^x|1-t|dt, xgt1),(x-1/2, xle1):} then

If f(x)= int_(-1)^(x)|t|dt ,x>=-1 then

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

Let g(x) = int_0^(x) f(t) dt , where 1/2 le f(t) < 1, t in [0,1] and 0 < f(t) le 1/2 "for" t in[1,2] . Then :

If f(x) = int_1^x (1)/(2 + t^4) dt ,then

If f(x)=int_(1)^(x)(ln t)/(1+t)dt, then

Let, f(x) = int_0^(x) (cos t)/t dt . Then x = (2n+1) pi/2 , f(x) has