绘制积分区域并证明两积分相等：非计算性论证要求

Got it, let's tackle this problem using just graphical reasoning and algebraic manipulation—no actual integral calculations, as requested.

First, let's map out the region we're dealing with:

• The curve defined by $y = \frac{1}{1+x^2}$ is a smooth, decreasing curve in the first quadrant. It starts at the point $(0,1)$ (when $x=0$, $y=1$) and approaches the x-axis ($y=0$) as $x$ grows infinitely large.
• The area this integral represents is bounded by:
  • The curve $y = \frac{1}{1+x^2}$
  • The x-axis ($y=0$)
  • The y-axis ($x=0$)
  • It extends infinitely far to the right along the x-axis.
• As you noted, this area can be thought of as stacking infinitely thin vertical rectangles (each of width $dx$) from $x=0$ all the way out to $x \to \infty$. Each rectangle's height is given by the curve's value at that $x$, $\frac{1}{1+x^2}$.

We can connect the two integrals by switching our perspective from vertical slices to horizontal slices, or using a simple algebraic substitution. Let's cover both angles:

Algebraic Argument #

Start with the curve's equation from the first integral: $y = \frac{1}{1+x^2}$. Let's rearrange this to solve for $x$ in terms of $y$ (since we want to shift variables from $x$ to $y$):

1. Multiply both sides by $1+x^2$: $y(1+x^2) = 1$
2. Expand and rearrange to isolate the $x^2$ term: $yx^2 = 1 - y$
3. Divide both sides by $y$ (we can do this because $y > 0$ everywhere in our region): $x^2 = \frac{1-y}{y}$
4. Take the positive square root (since we're only looking at $x \geq 0$ in the original integral): $x = \sqrt{\frac{1-y}{y}}$

Now, think about what this means for the area. The original integral sums up vertical slices ($y \cdot dx$) across $x$ from 0 to $\infty$. If we instead sum up horizontal slices ($x \cdot dy$), we need to find the range of $y$ values:

• When $x=0$, $y=1$; as $x \to \infty$, $y \to 0$. So $y$ ranges from 0 to 1.
• For each $y$ in that range, the horizontal width of our region is exactly the $x$ value we solved for: $\sqrt{\frac{1-y}{y}}$.

Summing these horizontal slices gives us the integral $\int_0^1 \sqrt{\frac{1-y}{y}} dy$—and since we're measuring the exact same area, this must equal the original integral.

Graphical Argument #

If you've already plotted the curve $y = \frac{1}{1+x^2}$:

• Imagine you've been counting the area by stacking rectangles standing upright (vertical slices) from left to right.
• Now, flip your perspective: instead of standing rectangles, lay them on their side (horizontal slices). Each horizontal slice at height $y$ stretches from the y-axis ($x=0$) to the curve at that height, which is $x = \sqrt{\frac{1-y}{y}}$.
• You'll be stacking these slices from the bottom of the region ($y=0$) up to the top ($y=1$). The total area from this horizontal sweep is exactly the second integral—and it's the same area as the vertical sweep, so the two integrals are equal.

内容的提问来源于stack exchange，提问作者chinny