text: # Fundamental theorem of calculus
  
  ## Main Question or Discussion Point
  
  There are two theorems:
  
  The fundamental theorem of calculus: $$\int_{a}^{b}F'(x) = F(b) - F(a)$$
  
  And the theorem that states if f is continious on [a,b]and g:[a,b]->R is defined by
  g(x) = $$\int_{a}^{x}f(t) dt$$, then g is differentiable on (a,b) and;
  
  g'(x) = $$\frac{d}{dx}(\int^{x}_{a}f(t)) = f(x)$$
  
  "There’s one question that asks if it is possible to readily show that each one implies the other, by a few lines of manipulation."
  
  Does anyone know how to do that?
  
  Let f be cts on [a,b], and let F(x) = $$\int f(x) dx$$ be called F, the antiderivative of f. Then
  
  $$\int_{a}^{b}f(x) dx = F(b) - F(a)$$
  
  By theorem $$\frac{d}{dx}(\int^{x}_{a}f(t)) = f(x)$$, we know $$\int_{a}^{x}f(t) dt = F(b) - F(a)$$ & F(x) has the same derivative, f(x). Hence there will be a constant c such that $$\int_{a}^{x}f(t)dt = F(x)+c$$ When x = a, we get
  
  F(a) + c = $$\int_{a}^{a}f(t)dt = 0$$ so c = -F(a)
  
  Hence, $$\int_{a}^{x}f(t)dt$$ = F(x)-F(a)
  When x = b this yields
  
  $$\int_{a}^{b}f(t)dt = F(b)-F(a)$$
  
  I'm not sure if this really answers the question though.
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