If `y=int_x^(x^2)t^2dt`, then the equation of tangent at `x=1` is

If y = int_(u(x))^(v(x))f(t) dt , let us define (dy)/(dx) in a different manner as (dy)/(dx) = v'(x) f^(2)(v(x)) - u'(x) f^(2)(u(x)) alnd the equation of the tangent at (a,b) as y -b = (dy/dx)_("(a,b)") (x-a) If y = int_(x)^(x^(2)) t^(2)dt , then equation of tangent at x = 1 is

If y =int _(u(x))^(v(x))f(t) dt , let us define (dy)/(dx) in a different manner as (dy)/(dx) = v'(x) f^(2)(v(x)) - u'(x) f^(2)(u(x)) alnd the equation of the tangent at (a,b) as y -b = (dy/dx)_((a,b)) (x-a) If F(x) = int_(1)^(x)e^(t^(2)//2)(1-t^(2))dt , then d/(dx) F(x) at x = 1 is

If y = underset(u(x))overset(v(x))intf(t) dt , let us define (dy)/(dx) in a different manner as (dy)/(dx) = v'(x) f^(2)(v(x)) - u'(x) f^(2)(u(x)) alnd the equation of the tangent at (a,b) as y -b = (dy/dx)_((a,b)) (x-a) If F(x) = underset(1)overset(x)inte^(t^(2)//2)(1-t^(2))dt , then d/(dx) F(x) at x = 1 is

If y=int_(0)^(x)(t^(2))/(sqrt(t^(2)+1))dt then rate of change of y with respect to x when x=1 is

Let y= int_(u(x))^(y(x)) f (t) dt, let us define (dy)/(dx) as (dy)/(dx)=v'(x) f^(2) (v(x)) - u' (x) f^(2) (u(x)) and the equation of the tangent at (a,b) and y-b=((dy)/(dx))(a,b) (x-a) . If f(x)=int _(x)^(x^2) t^(2) dt " then" (d)/(dx) f(x) at x = 1 is

If int_2^xg(t)dt=(x^2)/2+int_x^2t^2g(t)dt , then equation g(x)=lambda has 2 solution if |lambda|+ 1/2

If int_0^x(f(t))dt=x+int_x^1(t^2.f(t))dt+pi/4-1 , then the value of the integral int_-1^1(f(x))dx is equal to

f(x)=int_(0)^( pi)f(t)dt=x+int_(x)^(1)tf(t)dt, then the value of f(1) is (1)/(2)

If int_(0)^(x) f ( t) dt = x + int_(x)^(1) tf (t) dt , then the value of f(1) is

The slope of the tangent to the curve y=int_(0)^(x)(1)/(1+t^3)dt at the point, where x=1 is