If `f(y)=e^(y),g(y)=y,ygt0` and `F(t)=int_(0)^(1)f(t-y)g(y)dt` then

If f(y)=e^y,g(y)=y,y>0, and F(t)=int_0^t f(t-y)g(y) dy , then

If f(y)=e^y ,g(y) = y>0,a n d \ F(t)=int_0^t f(t-y)g(y)dy ,then

If int_(0)^(x)f(t)dt = x^(2)-int_(0)^(x^(2))(f(t))/(t)dt then find f(1) .

If int_(0)^(x)f(t)dt=e^(x)-ae^(2x)int_(0)^(1)f(t)e^(-t)dt , then

If f(x-y)=f(x).g(y)-f(y).g(x) and g(x-y)=g(x).g(y)+f(x).f(y) for all x in R . If right handed derivative at x=0 exists for f(x) find the derivative of g(x) at x =0

If f(x-y)=f(x).g(y)-f(y).g(x) and g(x-y)=g(x).g(y)+f(x).f(y) for all x in R . If right handed derivative at x=0 exists for f(x) find the derivative of g(x) at x =0

Suppose f(x) and g(x) are two continuous functions defined for 0<=x<=1 .Given, f(x)=int_0^1 e^(x+1) .f(t) dt and g(x)=int_0^1 e^(x+1) *g(t) dt+x The value of f( 1) equals

IF f(x+f(y))=f(x)+y AA x, y in R and f(0)=1 , then int_(0)^(10)f(10-x)dx is equal to

Given a function 'g' continous everywhere such that int _(0) ^(1) g (t ) dt =2 and g (1)=5. If f (x ) =1/2 int _(0) ^(x) (x-t) ^(2)g (t)dt, then the vlaue of f''(1)-f'(1) is:

Given a function g, continous everywhere such that g (1)=5 and int _(0)^(1) g (t) dt =2. If f (x) =1/2 int _(0) ^(x) (x -t)^(2) g (t) dt, then find the value of f '(1)+f''(1).