Whenever I take a definite integral in aim to calculate the area bound between two functions, what is the meaning of a negative result? Does it simly mean that the said area is under the the x - axis, in the negative domain of the axis?

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    $\begingroup$ Yes, this is simply that. Definite integrals compute the signed area between the curve and the $x$ axis. So, the signed area can be positive, negative or zero. For example, integrate $\sin(x)$ between $0$ and $\pi$, $\pi$ and $2\pi$,$0$ and $2\pi$. $\endgroup$ Jul 19, 2014 at 8:05
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    $\begingroup$ Note that $\int_1^0 1\, dx=-1$, so integrating a positive function in the wrong direction also yields a negative result. $\endgroup$
    – Ragnar
    Jul 19, 2014 at 10:44
  • $\begingroup$ take a look here math.stackexchange.com/questions/1316529/… $\endgroup$
    – user
    Dec 3, 2017 at 20:23
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    $\begingroup$ @Trey what is the reasoning for the bounty for official sources? This may help us understand what you're going after. $\endgroup$
    – rb612
    Dec 7, 2017 at 18:57

3 Answers 3


Not always.

The definite integral of a positive function (that is a function over the $x$-axis) gives a positive area. This is, $$\int_a^b f(x) dx \geq 0$$ for a function such that $f(x) \geq 0$ when $a<x<b$.

On the other hand, the definite integral of a negative function (that is a function under the $x$-axis) gives a negative area. This is $$\int_a^b f(x) dx \leq 0$$ for a function such that $f(x) \leq 0$ when $a<x<b$.

Now, the problem comes when you have a function that goes for a while over the $x$-axis, and for another while under it. For example, let's assume that goes over the axis on the interval $[a,c]$ and under it on the interval $[c,b]$. That is $$ f(x)\geq 0 \text{ when } a<x<c, $$ $$ f(x)\leq 0 \text{ when } c<x<b. $$ Then the integral can be split in two as $$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx = A_1 + A_2,$$ where $A_1\geq 0$ and $A_2 \leq 0$.

Then, depending on which of the two areas $A_1$ and $A_2$ is "bigger" the result of the definite integral will be positive or negative. That is, $$ \text{If } |A_1|\geq|A_2| \text { then } \int_a^b f(x) dx \geq 0 ,$$ $$ \text{If } |A_1|\leq|A_2| \text { then } \int_a^b f(x) dx \leq 0 .$$

To sum it up, a negative definite integral means that there is "more area" under the $x$-axis than over it.


Let us begin with something very important: Integration is wonderful! No, let us discover some of its wonderfulness. At first, as stated and as it is known, the definite integal of a Riemman-integrable positive function $f:[a,b]\to\mathbb{R}$ over an interval $[a,b]$ represents the area enclosed between the curve $f$ defines on the plane, te $x'x$ axis and the lines $x=a$ and $x=b$. which results to the fact that, if $\displaystyle\int_a^bf(x)dx$ is negative, then, at least a "large" enough part of the abovementioned area lies under the $x'x$ axis. All this has been said up to now.

However, as integrating is - in a way - a more generalised way to sum - uncountably many - numbers, it is expected that some of the summation properties should be "transferred" to integration. In deed, to define $f's$ average value, $\overline{f}$, we use the following integral: $$\overline{f}=\frac{1}{b-a}\int_a^bf(x)dx$$ So, we do exactly what we would do if we have to find the average value of the numbers $x_1,x_2,\dots,x_n$: $$\overline{x}=\frac{1}{n}\sum_{i=1}^nx_i$$ See, how $n$ is the "length" of the "interval" over which we are summing. Also note that the sign of $\overline{f}$ is independent of $\frac{1}{b-a}$, so, $$\overline{f}<0\Leftrightarrow\int_a^bf(x)dx<0$$ and $$\overline{f}>0\Leftrightarrow\int_a^bf(x)dx>0$$ So, if the integral of a function is negative then, it is expected, should one choose a point $y_0\in f([a,b])$ uniformly, that $y_0<0$ - of course, "expected" means "expected" and not "for sure".

A more interesting meaning of a negative integral comes from physics. Let $F:[a,b]\to\mathbb{R}$ be a function with $F(x)$ being the force applied to a mass when it is at position $x$. So, if $F$ was a constant over $[a,b]$, then, as we know from school physics, it work is given by the formula $$W_F=F\cdot(b-a)$$ Now, imagine that $F$ is not constant over $[a,b]$. So, let us "cut" $[a,b]$ to elementary small pieces of length $dx$ over which $F$ is "almost" constant - of course this is a rather intuitive approach, however it is very useful to explain the idea of the formula that will be given below. Then we have that, on these "pieces" the work of $F$ is given again by the formula $W_F=F\cdot dx$. Now, summing all these "partial works", we get that: $$W_F=\sum F\cdot dx$$ It is clear that the, mathematically, correct formula for this is - imagine that we are now letting $dx\to0$: $$W_F=\int_a^b F(x)dx$$ So, thinking of every function $F$ as the force applied to a mass while it is moving through $[a,b]$ gives a clear interpretation to any integral, positive or negative; it is the work of $F$ during that movement. So, that integral being negative, it means that we make more effort to "stop" this mass, than to support its movement.

I hope these were helpful! :)

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    $\begingroup$ the physics example is astonishing! $\endgroup$
    – Trey
    Dec 7, 2017 at 20:22
  • $\begingroup$ I know. I was also very impressed when first met this approach of finding a force's work over an interval. In a similar way, you can find many interesting formulas, as, for instance, for the length of a curve/function over an interval etc. $\endgroup$ Dec 7, 2017 at 20:52

This area thing is for kids. What you have to grasp is the general idea of an integral.

We are given a solid $B\subset{\mathbb R}^n$, $n\geq1$, and a function $f:\>B\to{\mathbb R}$ modeling some variable (geometrical, physical, etc.) intensity on $B$. The integral $\int_B f(x)\>{\rm d}(x)$ then encaptures the total impact caused by $f$ on this $B$ in terms of a single number. Thinking about this idea, and the properties this total impact should naturally have, brings us to the following setup: Partition $B$ into a large number of tiny subdomains $B_k$ $(1\leq k\leq N)$ and choose a sample point $x_k$ in each $B_k$. Then the quantity we have in mind is approximately given by $$\sum_{k=1}^N f(x_k)\>{\rm vol}(B_k)\ ,$$ so that we should define $$\int_B f(x)\>{\rm d}(x):=\lim_{\ldots}\sum_{k=1}^N f(x_k)\>{\rm vol}(B_k)\ ,$$ whereby the $\ldots$ under the $\lim$ will have to be filled with lots of legalese.

In the simple case of an $f:\>[a,b]\to{\mathbb R}_{>0}$ the function $f$ models an "area intensity" over the interval $[a,b]$, and the integral then is the "area under the curve".


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