Cambio de límites en integrales

Below is a common rigorous approach, first using the additivity property of integrals (which follows from the Fundamental Theorem of Calculus) and then connecting it to the sign change. I’ll outline two proofs—one that uses a very short “algebraic” argument once we trust a few fundamental properties of integrals, and one that uses a more “partition-based” argument that can be seen as a direct consequence of how Riemann sums work.

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1) Short Algebraic/FTOC-Based Proof

We start from one of the most important properties of definite integrals:

Additivity of integrals: If $a < b < c$, then

$$\int_a^b f(x)\,dx + \int_b^c f(x)\,dx \;=\; \int_a^c f(x)\,dx.$$

Now observe what happens if in that property we set $c = a$. We then have:

$$\int_a^b f(x)\,dx \;+\; \int_b^a f(x)\,dx \;=\; \int_a^a f(x)\,dx.$$

But $\int_a^a f(x)\,dx = 0$ (the integral over an interval of zero length is zero). Hence,

$$\int_a^b f(x)\,dx + \int_b^a f(x)\,dx \;=\; 0.$$

Rearranging gives:

$$\int_b^a f(x)\,dx \;=\; -\,\int_a^b f(x)\,dx.$$

That’s it—we have shown algebraically that reversing the limits multiplies the integral by $-1$.

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2) Partition/Riemann-Sum Argument

If we prefer something that looks closer to the definition via Riemann sums, we can argue as follows. Let $P = \{x_0, x_1, \dots, x_n\}$ be a partition of the interval $[a,b]$ where $a = x_0 < x_1 < \cdots < x_n = b$. For a Riemann sum, we pick sample points $\xi_i \in [x_{i-1}, x_i]$ and form

$$S_P \;=\; \sum_{i=1}^n f(\xi_i)\,\bigl(x_i - x_{i-1}\bigr).$$

As we refine the partition ($n \to \infty$, mesh size $\to 0$), $S_P \to \int_a^b f(x)\,dx$.

Now consider the interval $[b,a]$ (i.e., “backwards”). A partition of $[b,a]$ in the usual sense is $\{b = x_n > x_{n-1} > \dots > x_0 = a\}$. However, if you insist on always writing partitions in increasing order, you can view a partition of $[b,a]$ simply by writing the same points in reverse:

$$P' \;=\; \{a = x_0 < x_1 < \dots < x_n = b\},$$

but to traverse from $b$ down to $a$, each sub-interval “goes backward.” Concretely,

• The sub-interval from $x_n$ down to $x_{n-1}$ has length $(x_{n-1} - x_n)$, which is negative since $x_{n-1} < x_n$.
• Each term in the Riemann sum for $\int_b^a$ effectively has a negative $\Delta x$ (because we are going from a larger number to a smaller number).

Hence, if you wrote the Riemann sum in the same left-to-right order but conceptually are “going backward,” each term in the sum would gain a factor of $-1$. In the limit, this negative factor persists:

$$\int_b^a f(x)\,dx \;=\; \lim_{\|\Delta\|\to 0} \sum_{i=1}^n f(\eta_i)\,\bigl(x_{i-1} - x_i\bigr) \;=\; - \lim_{\|\Delta\|\to 0} \sum_{i=1}^n f(\eta_i)\,\bigl(x_i - x_{i-1}\bigr) \;=\; -\,\int_a^b f(x)\,dx.$$

This again shows, but now in a partition-based manner, that reversing the limits negates the integral.

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Why “Area” Intuition is Tricky

Your question mentions being confused because “the area under the curve from $a$ to $b$ is the same as from $b$ to $a$.” It’s true that geometrically, the region in the plane spanned by $x$ from $a$ to $b$ is the same as the region spanned by $x$ from $b$ to $a$. However, in integral calculus, we attach an orientation or direction to the integral—so it’s more akin to signed area. Traversing an interval left-to-right is considered the positive direction, whereas right-to-left is negative.

In short, definite integrals are not purely “the area under a curve” in the everyday sense—they are signed. This sign convention is consistent with many deeper theorems in analysis and also underlies the idea that $\int_a^b f'(x)\,dx = f(b)-f(a)$, which likewise changes sign if you swap $a$ and $b$.

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Summary

1. Using additivity and the fact $\int_a^a f(x)\,dx = 0$, we get:

   $$\int_a^b f(x)\,dx + \int_b^a f(x)\,dx = 0,$$

   which directly implies

   $$\int_b^a f(x)\,dx = -\,\int_a^b f(x)\,dx.$$

2. Using the definition via Riemann sums, one sees that reversing the interval introduces a negative increment $(x_{i-1}-x_i)$ instead of $(x_i - x_{i-1})$. This leads the limit to differ by an overall factor of $-1$.

These proofs show mathematically perfectly that reversing the limits of a definite integral multiplies its value by $-1$.

아래 설명은 “왜 적분을 정의할 때 구간 길이를 $|x_i - x_{i-1}|$로 안 쓰고 $(x_i - x_{i-1})$로 쓰는지”, 그리고 “왜 적분 구간을 뒤집으면 부호가 바뀌는지”를 조금 더 ‘근본적인’ 관점에서 다룹니다.

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1) 적분은 ‘순수 면적’이 아니라 ‘방향(부호) 있는 누적량’을 다룬다

먼저 우리가 흔히 $\int_a^b f(x)\,dx$를 “$f(x)$와 $x$-축이 만드는 면적”이라고 부르기는 하지만, 사실 ‘적분’은 방향(Orientation)이 들어간 누적량으로 정의됩니다.

• 구간 $[a,b]$를 오른쪽(증가 방향)으로 진행한다고 약속하면, $\int_a^b$는 ‘양(+)의 방향의 누적’이 되고,
• $[b,a]$처럼 왼쪽(감소 방향)으로 진행하면, ‘음(-)의 방향의 누적’이 됩니다.

즉, 적분은 “단순 면적”이라기보다는 **“(방향성을 가진) 면적 또는 누적량”**으로 정의되어 있기 때문에, 구간을 뒤집으면 부호가 바뀌는 것은 ‘정의’ 차원에서 자연스럽습니다.

만약 $\int_a^b$를 “면적의 절댓값”처럼 정의했다면, $\int_b^a$를 해도 똑같은 (양수) 값이 나오겠죠. 그러나 그렇게 하면 아래에서 볼 ‘기본정리(미적분학의 기본정리)’ 같은 핵심 성질들이 깨져버립니다.

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2) 미적분학의 기본정리와의 정합성

2.1 기본정리가 말하는 것

미적분학의 기본정리는

$$\int_a^b f'(x)\,dx \;=\; f(b) - f(a)$$

를 말합니다.

• 만약 적분을 $\int_a^b |x_i - x_{i-1}|\cdot f(\xi_i)$ 같은 식으로 (절댓값을 취해서) 정의해 버리면,
• 오른쪽에서 왼쪽으로 갈 때도 (부호 없이) 면적만 누적하게 되므로, $f(b)-f(a)$ 같은 “양 끝점의 함수값 차이” 공식을 그대로 만족할 수가 없습니다.

즉, 적분 구간이 $a\to b$인지 $b\to a$인지에 따라 결과가 달라져야 기본정리에서 말하는

$$\int_a^b f'(x)\,dx = f(b)-f(a), \quad \int_b^a f'(x)\,dx = f(a)-f(b) = -(f(b)-f(a))$$

가 일관성 있게 유지됩니다.

2.2 $F(b) - F(a)$를 통해 보는 “구간 뒤집기”

기본정리에 의해, 어떤 $F'(x)=f(x)$인 원시함수$F$가 있을 때,

$$\int_a^b f(x)\,dx \;=\; F(b) - F(a).$$

그렇다면 $\int_b^a f(x)\,dx$를 계산해 보면,

$$\int_b^a f(x)\,dx = F(a) - F(b) = -\bigl[F(b) - F(a)\bigr].$$

여기서 보이듯이, 적분 구간을 뒤집으면 $\int_a^b$ 값에 음(-)부호가 붙습니다.
이것이 적분을 “부호 있는 누적”으로 정의했을 때, 가장 깔끔하고 연속적으로 설명됩니다.

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3) 왜 $(x_i - x_{i-1})$로 정의할까? (절댓값 말고)

3.1 ‘방향’ 개념을 살리기 위해

적분은 “누적”을 표현합니다. 그런데 누적에도 방향이 있습니다. 예를 들어 물리에서,

• $x$가 증가하는 방향(왼쪽에서 오른쪽)으로 힘을 가하면, 일을 양(+)으로 계산하고,
• $x$가 감소하는 방향(오른쪽에서 왼쪽)으로 힘을 가하면, 일을 음(-)으로 계산하죠.

이처럼 방향까지 표현하려면, 부분구간 길이를 $|x_i - x_{i-1}|$로 처리해버리면 안 되고, “(끝점 - 시작점)” 그대로 두어서 “양의 방향이냐, 음의 방향이냐”를 구분해야 합니다.

3.2 부분합(Riemann Sum)에서 보는 직관

리만합(Riemann sum)을 생각해봅시다.

$$\sum_{i=1}^n f(\xi_i)\,\bigl[x_i - x_{i-1}\bigr].$$

• $x_i > x_{i-1}$이면 $(x_i - x_{i-1}) > 0$ 이므로, ‘오른쪽 방향으로 간 구간’을 양의 기여로 누적.
• $x_i < x_{i-1}$이면 $(x_i - x_{i-1}) < 0$ 이므로, ‘왼쪽 방향으로 간 구간’을 음의 기여로 누적.

즉, 각각의 ‘미소(미세) 구간 $\Delta x$’에 대해 방향성이 반영되어야 하고, 이것이 최종적인 적분의 부호를 결정합니다. 순수 면적만 누적하고 싶다면 절댓값을 씌우면 되지만, 그러면 “오른쪽에서 왼쪽으로 가는 과정”이 모두 그냥 양수로 합쳐져버려, 위에서 말한 물리적 관점이나 기본정리 관점이 다 깨져버리게 됩니다.

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4) “왜 적분 구간을 바꾸면 부호가 반대가 되는가?”에 대한 좀 더 근본적인 요약

1. 정의 자체가 ‘방향(Orientation)’이 포함된 누적량이기 때문

   • 적분은 면적을 ‘부호 있게’ 센다는 아이디어에서 출발한다.
   • $(x_i - x_{i-1})$가 양수면 오른쪽 이동(+) 기여, 음수면 왼쪽 이동(-) 기여로 계산.

2. 미적분학의 기본정리와 일관성을 맞추기 위해

   • $\int_a^b f'(x)\,dx = f(b)-f(a)$라는 공식이 자연스럽게 유지되려면,
   • $\int_b^a$를 했을 때 $\int_a^b$의 ‘음수 버전’이 되어야 한다.

3. 절댓값으로 정의하면, 이 모든 성질이 망가진다

   • 절댓값을 쓰면 “면적”만 나오므로, 방향성이 사라지고, $f(b)-f(a)$와의 직접적 연결(기본정리)도 깨진다.

정리하자면, 적분은 “왼쪽에서 오른쪽으로 (또는 작은 $x$에서 큰 $x$로) 갈 때”를 **‘기준’**으로 하여 양(+) 방향으로 삼고, 그 반대로 갈 때는 음(-) 방향으로 잡는 **“부호가 있는 누적량”**으로 정의됩니다.
이 방식으로 정의해야 미분과 적분이 서로 딱 들어맞고, 우리가 원하는 (물리, 공학, 수학) 이론이 깔끔하게 돌아갑니다.

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결론

• “적분” = “(방향이 있는) 면적 누적”으로 정의
• 구간 뒤집기 → 방향 뒤집기 → 부호 반전
• 이는 면적 자체를 ‘그냥 양수로만 모으는 것’과 달리, 방향 정보를 포함하기 위해 그렇게 정의한 것.
• 덕분에 미적분학의 기본정리를 비롯한 여러 중요한 성질이 성립.

이 모든 것이 **“왜 적분을 그렇게(방향을 살려) 정의하느냐?”**에 대한 가장 근본적인 이유입니다.
“방향이 붙어야 미분과 적분이 서로 맞물린다.”라고 이해하시면 좋겠습니다.

Below is a more “from first principles” explanation of why reversing the limits of integration introduces a minus sign, but without just saying “memorize it.” We will use the idea of constructing a function $S(x)$ (which will turn out to be an antiderivative of $f$) and see how its values naturally give us the integral from one point to another. From that construction, we will see very explicitly why swapping the endpoints reverses the sign.

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1. Defining $S(x)$ and Connecting It to $\int$

Let’s define a function $S(x)$ by picking some fixed point $c$ (think of it as a “reference point”), and then writing

$$S(x) \;=\; \int_{c}^{\,x} f(t)\,dt.$$

This means: $S(x)$ is the “accumulated area” (with sign!) of $f$ as $t$ goes from $c$ up to $x$.

1. If $x > c$, the integral $\int_c^x f(t)\,dt$ accumulates “positive area” (or negative, if $f$ dips below the axis, but the direction is $c \to x$).
2. If $x < c$, then $\int_c^x f(t)\,dt$ accumulates “area” in the $\textit{reverse}$ direction, and so $S(x)$ might be negative relative to $S(c)$.

In any case, $S(x)$ is well-defined once $f$ is integrable.

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2. The Fundamental Theorem of Calculus in Terms of $S(x)$

A key result (the Fundamental Theorem of Calculus, Part 1) says that if $f$ is continuous, then $S'(x) = f(x)$. In other words:

$$\frac{d}{dx} \Bigl(\int_c^x f(t)\,dt\Bigr) \;=\; f(x).$$

This theorem is profound because it links differentiation and integration. But more importantly for our question, it tells us:

$\int_a^b f(x)\,dx = S(b) - S(a)$,
since $\int_a^b f(x)\,dx$ can be written as $\int_c^b f(t)\,dt - \int_c^a f(t)\,dt = S(b) - S(a).$

Hence, we identify $\int_a^b f(x)\,dx = S(b) - S(a)$.

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3. What Happens If We Swap $a$ and $b$?

Now let’s look at $\int_b^a f(x)\,dx$. By exactly the same reasoning,

$$\int_b^a f(x)\,dx \;=\; S(a) \;-\; S(b).$$

But notice:

$$S(a) - S(b) \;=\; -\bigl[S(b) - S(a)\bigr].$$

Since $S(b) - S(a) = \int_a^b f(x)\,dx$, we immediately see

$$\int_b^a f(x)\,dx = -\,\int_a^b f(x)\,dx.$$

That’s it: the minus sign emerges directly from the fact that $\int_a^b f(x)\,dx = S(b) - S(a)$, so switching $a$ and $b$ gives $S(a) - S(b)$. This is a purely algebraic consequence once we define $S(x)$ that way.

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4. Why Not Just “Sum the Sticks” in Either Order?

You asked: “Aren’t we just summing little rectangle areas (Riemann sums)? Why does summing them from right to left flip the sign?”

4.1 Riemann sums are oriented sums

When we define a Riemann sum on $[a,b]$:

$$\sum_{i=1}^n f(\xi_i)\,\bigl(x_i - x_{i-1}\bigr),$$

we do not take the absolute value $\bigl|x_i - x_{i-1}\bigr|$. Instead, we literally use $(x_i - x_{i-1})$. If $x_i > x_{i-1}$, that difference is positive and contributes positively; if we were to swap the order and go from $b$ to $a$, each little piece becomes $(x_{i-1} - x_i)$, which is negative of $(x_i - x_{i-1})$.

Thus, when we sum up from right to left (that is, if the partition is traversed in decreasing order), each sub-interval’s width is negative, and so the total sum becomes the negative of what it was when going left to right.

4.2 “But why define it that way?”

Because we want consistency with the Fundamental Theorem of Calculus. That theorem yields

$$\int_a^b f'(x)\,dx \;=\; f(b) - f(a).$$

If we had defined the integral using absolute values of widths, then $\int_b^a f'(x)\,dx$ would no longer come out to $f(a) - f(b)$. We would lose the beautiful statement that $\int_a^b f'(x)\,dx$ is exactly the net change $f(b)-f(a)$.

Put differently, to match the idea of “net change” (rather than “absolute accumulation”), we need the orientation to matter. And that orientation is exactly why $\Delta x_i = (x_i - x_{i-1})$ instead of $\lvert x_i - x_{i-1}\rvert$.

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5. Geometric vs. Algebraic/Net-Change View

5.1 Geometric interpretation

• Sometimes we say “$\int_a^b f(x)\,dx$ is the signed area under $f$.”
• “Signed” here means that if we move left-to-right, that area is taken as positive if $f$ is above the $x$-axis, negative if $f$ is below, and if we literally reverse the direction (i.e., go from $b$ down to $a$), the entire orientation of that shape is reversed, hence the sign flips.

5.2 Algebraic/Net-change interpretation

• $\int_a^b f(x)\,dx$ is the “net change” of the antiderivative $F$ of $f$, because $F'(x) = f(x)$.
• That net change is $F(b) - F(a)$. Reversing the interval naturally becomes $F(a) - F(b)$, i.e. the negative.

Either viewpoint leads us to the same conclusion that reversing endpoints introduces a minus sign. This is not a random convention—it is forced if we want the integral to measure net change in a consistent way with derivatives.

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6. Summary

1. We define a function $S(x) = \int_{c}^x f(t)\,dt$.
2. Fundamental Theorem of Calculus says $S'(x) = f(x)$.
3. The definite integral $\int_a^b f(x)\,dx$ is then $S(b) - S(a)$.
4. If we flip $a$ and $b$, we get $S(a) - S(b)$, which is exactly the negative.

This derivation shows that the minus sign is not arbitrary nor something to “just memorize.” It is an inevitable outcome of how $S(x)$ is constructed to be an antiderivative of $f$, combined with the fundamental theorem linking integrals and derivatives. We do not use absolute differences in the Riemann sum because that would break the key property that $\int_a^b f'(x)\,dx = f(b)-f(a)$.

Hence, the most fundamental reason is that the integral represents net (oriented) change rather than just “the size of stacked rectangles.” That orientation is built in mathematically through the difference $(x_i - x_{i-1})$, and it precisely explains why reversing the limits of integration introduces a minus sign.