`Iff(y)=e^y ,g(y)=y;y>0,a n dF(t)=int_0^t f(t-y)g(y) dy,t h e n` a) `F(t)=e^t-(1+t)` b) `F(t)=t e^t` c) `F(t)=t e^(-t)` (d) `F(t)=1-e^t(1+t)`

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

If int_0^1 (e^t dt)/(t+1)=a, then evaluate int_(b-1)^b (e^t dt)/(t-b-1)

Ifx=int_0^y(dt)/(sqrt(1+9t^2))a n d(d^2y)/(dx^2)=a y ,t h e nfin da

Let f(x)=int_(0)^(x)(e^(t))/(t)dt(xgt0), then e^(-a)[f(x+1)-f(1+a)]=

Let g (x,y)=x^2-yx + sin (x+y), x (t) = e^(3t) , y(t) = t^2, t in R. Find (dg)/(dt) .

If x=t^2,y=t^3,t h e n(d^2y)/(dx^2)= (a) 3/2 (b) 3/((4t)) (c) 3/(2(t)) (d) (3t)/2

Integrate the functions e^(t)((1)/(t)-(1)/(t^(2)))

Ifx=int_c^(sint)sin^(-1)z dz ,y=int_k^(sqrt(t))(sinz^2)/zdz ,t h e n(dy)/(dx)i se q u a lto (tant)/(2t) (b) (tant)/(t^2) (c) tan t/(2t^2) (d) (t a tt^2)/(2t^2)

If g(x,y) = 3x^(2) - 5y + 2y^(2), x(t) =e^(t) and y(t) = cos t , then (dg)/(dt) is equal to