2
$\begingroup$

The slope field for $F^{'}(x) = e^{-x^{2}}$ is shown here with a particular solution $F(0)=0$ superimposed. Using calculator, find

$$ \lim_{x \rightarrow \infty} F(x)$$ to $3$ decimal places.

Answer choices were

(A) $0.886$

(B) $0.987$

(C) $1.000$

(D) $1.414$

(E) $\infty$

The book says that answer is A, but I want to understand this.

enter image description here

2 Answers 2

1

It is well known that

$$\int_{0}^{\infty} e^{-x^2} \text{ dx} = \frac{\sqrt{\pi}}{2} = 0.8862269\dots$$

For instance see here: http://en.wikipedia.org/wiki/Gaussian_integral

or

Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \dfrac{\sqrt \pi}{2}$

So that explains why the answer is A.

Were you looking for some other explanation?

  • 0
    I was looking for an explanation that uses the superimposed image. There should be a reason why they gave that. (probably) I kind of knew about Error function by doing web search.2012-03-10
  • 0
    @Sam: I have no clue what the image means :-) Anyway, if you are asking us to guess the solution they _want_, I guess it is impossible unless you give more context.2012-03-10
  • 0
    Does $F^{'}(x) = e^{-x^{2}}$ mean $F(x) = \int_{0}^{x} e^{-t^{2}} \text{dt}$ and therefore the expected limit is $\int_{0}^{\infty} e^{-x^2} \text{ dx}$ ?2012-03-10
  • 3
    @Sam I surmise they want you to (using the calculator) put your crosshair on the right edge of the solution curve and record the $y$-coordinate.2012-03-10
  • 0
    @Sam: Yes, that and $F(0) = 0$.2012-03-10
  • 0
    @DavidMitra: Post this as an answer?2012-08-14
1

I am also curious what calculator is here for. We just need a crude idea of what $\int_0^\infty e^{-x^2}\,dx$ is. Note that $\int_a^{\infty}e^{-x^2}\,dx\le e^{-a^2}\int_0^\infty e^{-2at}\,dt=e^{-a^2}/(2a)$ and that $e^{-4}/4< 0.01$, so we really need only $$ \int_0^2e^{-x^2}\,dx\approx \frac 13(1+4e^{-1}+e^{-4})\approx (1+\frac 4{2.7}+\frac 1{7.29})/3\approx \frac 13+\frac 12+\frac 1{22}=0.83333\dots+0.04545\dots\approx 0.878\dots $$ Of course, Simpson is a slight underestimate here but for the multiple choice, it is perfectly suitable.

Of course, if you happen to know the exact value $\sqrt\pi/2$, the life gets completely trivial.