Final Thoughts — Question 1

Question 1 – Solution

Strategy. A definite integral defines a function. The Fundamental Theorem of Calculus can verify that function even when elementary antidifferentiation is unavailable.

Step 1: Verification and domain. The integrand is continuous for every real $t$, so the integral exists for every finite real $x$. The Fundamental Theorem gives $$F'(x)=e^{-x^2},\qquad F(0)=1+0=1.$$ Thus $\boxed{F\text{ is an exact solution on }\mathbb R.}$ Negative $x$ causes no problem: the integral is interpreted with reversed limits.

Step 2: Uniqueness without a general theorem. If $G$ solves the same IVP on an interval $I$ containing $0$, then $$(G-F)'=e^{-x^2}-e^{-x^2}=0.$$ A differentiable function with zero derivative on an interval is constant there. At $0$, $G-F=0$, so $\boxed{G(x)=F(x)\quad(x\in I).}$

Step 3: Strict integral bounds. For $0<t<1$, $$e^{-1}<e^{-t^2}<1.$$ These strict inequalities hold throughout an interval of positive length. Integrating from $0$ to $1$ therefore gives $$e^{-1}<\int_0^1 e^{-t^2}\,dt<1, \qquad \boxed{1+e^{-1}<F(1)<2.}$$

Step 4: What counts as a solution. The definite-integral expression specifies $F$ exactly, with no rounding. An elementary formula is a more restricted type of expression, assembled from familiar elementary functions. A decimal obtained by numerical integration is an approximation to a value of $F$. Failure to find an elementary expression does not negate the exact solution already verified.