Let's break this down step by step—your confusion about the integral result makes sense because there's a key mix-up in the integrand that we'll fix first.

1. Find the largest inscribed circle between ( y = e^{-x2} ) and ( y = -e^{-x2} ) #

Since the curves are symmetric across both the x-axis and y-axis, the largest circle must be centered at the origin (this is the only symmetric position that maximizes the radius without intersecting the curves). Let the circle's radius be ( r ), so its equation is ( x^2 + y^2 = r^2 ).

To find the tangent point between the circle and ( y = e^{-x2} ), we need two conditions:

• The function values match at the tangent point: ( e^{-x2} = y ) and ( x^2 + y^2 = r^2 )
• The slopes (derivatives) match at the tangent point:
  • Derivative of the curve: ( y' = -2x e^{-x2} )
  • Derivative of the circle: Differentiate ( x^2 + y^2 = r^2 ) to get ( 2x + 2y y' = 0 ), so ( y' = -\frac{x}{y} )

Set the derivatives equal (ignoring ( x=0 ), since that would give a circle of radius 1 which intersects the curves elsewhere):
[ -2x e^{-x2} = -\frac{x}{y} ]
Cancel out ( -x ), then substitute ( y = e^{-x2} ):
[ 2e^{-x2} = \frac{1}{e^{-x2}} \implies 2e^{-2x2} = 1 \implies e^{-2x2} = \frac{1}{2} ]
Solve for ( x^2 ): ( x^2 = \frac{\ln2}{2} ), and ( y = e^{-x2} = \frac{1}{\sqrt{2}} ).

Plug these into the circle's equation to find ( r^2 ):
[ r^2 = x^2 + y^2 = \frac{\ln2}{2} + \frac{1}{2} = \frac{1 + \ln2}{2} \approx 0.8466 ]

The area of the circle is:
[ S_{\text{circle}} = \pi r^2 = \frac{\pi(1 + \ln2)}{2} \approx 2.660 ]

2. Fix your integral mistake #

The integral you calculated ( \int_{-1}^1 2e^{-2x2}dx \approx 2.395 ) uses the wrong integrand. The vertical distance between ( y = e^{-x2} ) and ( y = -e^{-x2} ) is ( e^{-x2} - (-e^{-x2}) = 2e^{-x2} ), not ( 2e^{-2x2} ).

• For the area between the curves from ( x=-1 ) to ( x=1 ):
  [ \int_{-1}^1 2e^{-x2}dx = 4\int_0^1 e^{-x2}dx \approx 4 \times 0.7468 \approx 2.987 ]
• For the total area between the curves across the entire x-axis (since ( e^{-x2} \to 0 ) as ( x \to \pm\infty )):
  Using the Gaussian integral ( \int_{-\infty}^\infty e^{-x2}dx = \sqrt{\pi} ):
  [ S_{\text{total}} = \int_{-\infty}^\infty 2e^{-x2}dx = 2\sqrt{\pi} \approx 3.5449 ]

3. Calculate the area outside the circle #

• Across the entire plane:
  Subtract the circle's area from the total curve area:
  [ S_{\text{outside (total)}} = S_{\text{total}} - S_{\text{circle}} \approx 3.5449 - 2.660 \approx 0.885 ]
• Only between ( x=-1 ) and ( x=1 ):
  The circle spans ( x \in [-r, r] \approx [-0.92, 0.92] ), which is entirely within [-1,1]. So subtract the full circle area from the [-1,1] curve area:
  [ S_{\text{outside (-1,1)}} = \int_{-1}^1 2e^{-x2}dx - S_{\text{circle}} \approx 2.987 - 2.660 \approx 0.327 ]

Wrap-up #

Your initial doubt was well-founded—you used the wrong integrand in your integral. Correcting that gives you the accurate area values you need.

内容的提问来源于stack exchange，提问作者user517526