TI-84 integration algorithm is an adaptive Gauss-Kronrod 15-point rule, with apparently an absolute error threshold of $10^{-5}$ (TI-84 Plus and TI-84 Plus Silver Edition Guidebook, 2010, p. 41). When the absolute error estimate from the rule is greater than the threshold, the interval is subdivided and the integral recalculated over the intervals. The intervals are recursively subdivided until the error estimate is within the threshold.
For $a \le 892.26$, the initial error estimate is greater than the threshold,
and recursive subdivision is applied.
For $a \ge 892.27$, the initial error estimate is less than the threshold,
and no subdivision is applied.
Below is the equivalent computation done with Mathematica. For the OP's integral, to reproduce the integral, I had, surprisingly, to apply a common strategy used to cancel out odd functions when integrating over symmetric intervals:
$$\int_{-a}^{a} f(x) \; dx = \int_0^a {\left(f(x) + f(-x)\right)} \; dx$$
Here is the code:
{nodes, (* vector of nodes for the interval {0, 1} *)
wts, (* weights vector *)
errwts} = (* error weights vector *)
NIntegrate`GaussKronrodRuleData[7, MachinePrecision];
Length@nodes
(* 15 *)
The following shows that $10^{-5}$ is the threshold:
fvec = Exp[-Rescale[nodes, {0, 1}, {0, 892.26}]^2]; (* vector of function values *)
fvec.wts (2 * 892.26) (* Dot product of function values and nodes scaled by
interval length yields the integral *)
fvec.errwts (2 * 892.26) (* Scaled dot product with the error weights estimates the error *)
(*
0.0000100015 integral estimate
0.0000100015 error estimate
*)
fvec = Exp[-Rescale[nodes, {0, 1}, {0, 892.27}]^2];
fvec.wts (2 * 892.27)
fvec.errwts (2 * 892.27)
(*
9.99831*10^-6 integral estimate
9.99831*10^-6 error estimate
*)
The following calculates the integral and error estimate. The integral agrees with the OP's:
fvec = Exp[-Rescale[nodes, {0, 1}, {0, 1000}]^2];
fvec.wts (2000)
fvec.errwts (2000)
(*
2.71313*10^-7 integral estimate
2.71313*10^-7 error estimate
*)