solve for x: ∫ e^(-t^2) dt from 0 to x = 0.6?

Question

the antiderivitive of e^(-t^2) cannot be found analytically. I have a graphing calculator that can take numerical integrals; even so, I cannot think of any way to solve the problem without looking at a table of values.

Answer 1

I suspect that there is no analytical way to solve this. You might have to write a program to evaluate the integral numerically and the use a root-finding algorithm (such as the bisection method etc.) to get an adequate approximation to the value of x.

Answer 2

Do you know about Taylor series? If so,
expand e^(-t^2) about t=0, then integrate term-by-term:

e^(-t^2) = 1 - t^2 + (t^4)/2! - (t^6)/3! + -...

∫ e^(-t^2) dt = ∫ (1 - t^2 + (t^4)/2! - (t^6)/3! + -...) dt
= t - (t^3)/3 + (t^5)/10 - (t^7)/42 + - ...

At the lower limit, this is zero; at the upper limit, it's

0.6 - (0.6^3)/3 + (0.6^5)/10 - (0.6^7)/42 + ...
= 0.6 - .072 + 0.007776 - .0006665143 +...
≈ 0.5351

This is an alternating series with terms of decreasing magnitude, so the error incurred in truncating is less than the first term left off. The next term would be (0.6^9)/216≈0.000047 so the estimate above is correct to at least 3 decimal places (numerical integration gives 0.5351535)

Clearly, we wouldn't use this approach to estimate the integral if the upper limit were, say, 5 (even though the series does converge for all t -- you would need many, many terms, and unless you did "exact" arithmetic in the evaluation, the roundoff error would swamp any meaningful result).

Answer 3

How'd u type an integral sign?