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if $\int_0^1 f(x) \;\mathrm dx \,=\, 7$ and $\int_0^3 f(x) \;\mathrm dx = 4$, what is the value of $\int_1^3 f(x) \;\mathrm dx$?

In my opinion the value is 3 but it could be -3 ... how do I tell if it is 3 or -3?

From what I understand, the integral is an area, and area cant be negative — but then some questions I have seen do end up giving a negative answer. I'm confused!

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The definite integral gives area if the curve is above the $x$-axis. Area is nonnegative, but not so for the integral. If a curve is below the $x$-axis over $[a,b]$, $\int_a^b f(x)\,dx$ is negative; but its absolute value is the area bounded by the curve and $[a,b]$. So, to answer your question, it's $-3$. –  David Mitra Nov 23 '11 at 20:27
    
Hello nadal. I have reformatted your question in the hopes of making it clearer. Is this still the question you want to ask? –  Niel de Beaudrap Nov 23 '11 at 20:29
    
Integral coincides with the area under the graph if the function is nonnegative, but it is in general not true. If we have to stick to a geometric meaning anyhow, we may say that it refers to the signed area. –  sos440 Nov 23 '11 at 20:32
    
@niel yes thats the question –  nadal Nov 23 '11 at 20:37
    
If $a \le b$, then the area between the curves $y=f(x)$ and $y=g(x)$, from $x=a$ to $x=b$, is $\displaystyle\int_a^b|f(x)-g(x)|\,dx$. If you want to cut out a paper ornament that has shape the region between $y=\sin x$ and $y=-\sin x$, from $x=0$ to $x=2\pi$, you will need a positive amount of paper. –  André Nicolas Nov 23 '11 at 20:47

4 Answers 4

up vote 2 down vote accepted

The integral is $\int_1^3 f(x) dx= \int_0^3 f(x) dx -\int_0^1 f(x)dx=4 -7 =-3$.

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The following holds for definite integrals:

1) $\int_a^b f(x)\, dx =\int_a^c f(x)\,dx+\int_c^b f(x)\,dx\ $, for any numbers $a$, $b$, and $c$

and

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

In your setup, you could use $$\int_0^1 f(x)\,dx +\int_1^3f(x)\,dx =\int_0^3f(x)\,dx.$$

$$7+\int_1^3f(x)\,dx = 4\quad\Rightarrow\quad\int_1^3f(x)\,dx=4-7=-3.$$

The definite integral gives an area if the graph of the function you're integrating is above the $x$-axis:

If $f(x)\ge0 $ over $[a,b]$, then $\int_a^bf(x)\,dx$ is the area bounded by the graph of $f$, the $x$-axis, and the lines $x=a$, $x=b$.

Area is nonnegative, but not so for the integral. If the graph of the function you're integrating is below the $x$-axis over $[a,b]$, then $\int_a^b f(x)\,dx$ is negative; but its absolute value is the area bounded by the graph of $f$, the $x$-axis, and the lines $x=a$, $x=b$:

If $f(x)\le0 $ over $[a,b]$, then $\int_a^bf(x)\,dx$ is the negative of the area bounded by the graph of $f$, the $x$-axis, and the lines $x=a$, $x=b$.

In your example, the graph of $f$ would be above the $x$-axis (at least in part) over $[0,1]$ with "area" 7 ($\int_0^1 f(x)\thinspace dx=7$). You were told $\int_0^3 f(x)\,dx=4$. So over $[1,3]$, the graph would have to dip below the $x$-axis: assuming the graph of $f$ is always below the $x$-axis over $[1,3]$, there would have to be 3 units of area below the $x$-axis in order for the integral over $[0,3]$ to be 4, by rule 1) above.

In a nutshell: integrating over an interval over which $f(x)>0$ counts the area as positive, and integrating over an interval over which $f(x)<0$ counts the area as negative.

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When you do an integration, although it is often described as an area, that doesn't mean that it always corresponds literally to a geometric area of the sort you are used to.

Suppose you payed someone to do your plumbing: and you start paying them at \$5/hour, but the rate at which you pay them decreases continuously with time like the function $f(x) = 1/(x+5)^2$. You don't have any money to begin with, but the plumber will allow you to pay him later: so you will be in debt to him. If you express debt as a negative amount of money, how much money do you have if it takes him 3 hours to fix? You don't have to do any calculations to see that the value must be negative; you will owe the plumber something, which is a negative amount of money. And yet the result is expressible as an integral, of the form $$ \int_0^3 \frac{-1}{(x+5)^2} \;\mathrm dx $$ which corresponds to an area below the x-axis, but above the curve $y = -1/(x+5)^2$. Such areas below the x-axis and above a second curve always correspond to negative integrals.

If you like, you can think of these negative integrals as a generalization of area, which allows you to add negative areas as a way of removing area. But if that concept doesn't work for you, you can just remember that the interpretation as "just the area" isn't exact; it depends on being above the x-axis to be true (and if it's below the x-axis, the integral is the negative of the "geometric" area you're used to).

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It may help to understand why this happens by looking at the function $f(x)=-\frac{17}{3}x+\frac{59}{6}$. You can see a graph of the function here. If you integrate it, you'll find that it satisfies $\int_0^1f(x)dx=7$, $\int_0^3 f(x) dx=4$, and you can check that $\int_1^3 f(x) dx=-3$.

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