text: 1. This is true by Fundamental Theorem of Calculus.\n\nLet f be a continuous real function defined on a closed interval [a,b]. F(x) is an antiderivative of f(x) means that F(x) = Integrate f(t) dt from a to x. Suppose there is x and x+h inside [a,b], we have F(x) = Integrate f(t) dt from a to x AND F(x+h) = Integrate f(t) dt from a to x+h. The definition for dy/dx is such a limit on \n[F(x+h)-F(x)]/h in which h tends to zero. F(x+h)-F(x)= "Integrate f(t) dt from a to x+h" MINUS "Integrate f(t) dt from a to x" which is EQUAL TO "Integrate f(t) dt from x to x+h" According to mean value theorem, since f is continuos so is F, there exist a value c between [x,x+h] such that "Integrate f(t) dt from x to x+h" EQUALS "f(c) h". Then [F(x+h)-F(x)]/h = f(c) h/h = f(c). When u tends h to zero, since c is squeezed between [x,x+h], c tends to x (sandwich theorem). Hence dy/dx = f(x).\nTherefore, from y=F(x) is an antiderivative of f(x), you arrive at a conclusion dy/dx=f(x) which is equivalent to your question 1.\n\n2. This is also true by Fundamental Theorem of Calculus.(In fact this is a corollary(small truth following from the truth of a theorem, informally means small theorem)\n\nIf f(x) = F' (x) (ie dF(x)/dx). Therefore we know that limit h->0 for {[F(x+h)-F(x)]/h} is f(x). By mean value theorem, F'(c) = [F(x+h)-F(x)]/h for c in between [x,x+h]. {We assumed continuos function again, if not it doesnt work. In fact all your question means continuos function which is differential at all points within the function interval that we are interested in.} From f(x)= F'(x) we get f(c)=F'(c). Then you have F(x+h) = f(c)Xh + F(x). Since F(x+h) = f(c) h + F(x) for h tends to zero. Lets say we know that a certain initial value F(a). We want to find F(t) in the interval [a,x], then we can first start with F(a+h)=f(c) h + F(x) where c is between [a,a+h] and continue up till x. If we want to find F(x), it is just summation f(t) delta t, and if we limit delta x to zero, we get F(x) = Integrate f(t) dt from a to x  PLUS F(a), which is the antiderivative of f(x).\n\n3. x'' = k for k constant. Then x' = kt + C. x= k/2 * t^2 + Ct + D. Just do direct integration using statement 2.