For a function such as $f(x) = e^{x^{2}}$, how do I go about finding an anti-derivative for this that passes through a given point? Graphically, there should be no function that has derivative F'(x) = $e^{x^{2}}$, may I have some hints as to how I am able to tackle such a problem?

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    $\begingroup$ "Graphically, there should be no function that has derivative $F'(x)=e^{x^2}$" - Why not? I see no graphical issue with such a function, but if you elaborate on that point, it might help us understand what you're thinking and give a better answer. $\endgroup$ – Milo Brandt Jan 18 '16 at 6:22
  • $\begingroup$ Well, I was thinking of integrating over all values to find an indefinite integral, but after refreshing the definition from an answer below. I believe there is now if we set the bounds right as the area is finite. I guess what I want to do is a method to find a function whose derivative is $e^x^2$. $\endgroup$ – Mid Jan 18 '16 at 6:41
  • $\begingroup$ There is a function $F(x)$ such that $F'(x)=e^{x^2}$. It is given by $F(x)=\frac{1}{2} \sqrt{\pi }\, \text{erfi}(x)+C$ where appears the imaginary error function. $\endgroup$ – Claude Leibovici Jan 18 '16 at 8:32

Since $f$ is continuous on $\mathbb{R}$, it is Riemann integrable on $[c,x]$ for any $c \leq x$. By the fundamental theorem of calculus, $f: \mathbb{R} \to \mathbb{R}$ has an antiderivative $F: \mathbb{R} \to \mathbb{R}$ $$F(x) = \int_c^x f(t) \, dt$$ You can view this as a definition.

  • $\begingroup$ Oh does this involve F(X) - F(C) =$\int_{a}^{x} e^{t^2} dt$. So with a specific point (a,b) we can have F(a) = b so F(X) - b = $\int_{a}^{x} e^{t^2} dt$. But then I am just stuck with the original integral + a constant b? $\endgroup$ – Mid Jan 18 '16 at 6:32
  • $\begingroup$ It does answer the question, but a bit too simple? $\endgroup$ – Mid Jan 18 '16 at 6:55
  • $\begingroup$ I think so. Note that there are more than one antiderivatives. In my answer, the antiderivative is the one which passes through $(c,0)$. You can add a constant $b$ to the right hand side to make it passes through $(c,b)$. $\endgroup$ – Empiricist Jan 18 '16 at 6:58
  • $\begingroup$ Ah, so it would just be $F(X) = \int_{a}^{x} e^{t^2} dt + b$ would be a proper solution for a function with derivative $e^{t^2}$ that passes through point (a,b)? $\endgroup$ – Mid Jan 18 '16 at 7:03
  • $\begingroup$ Yes, I think so. :) $\endgroup$ – Empiricist Jan 18 '16 at 8:53

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