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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 correct statement is that the definite integral represents the *net signed area*. That means that area above the $x$-axis that is "traversed" from left to right counts as positive, area above the $x$-axis traversed from right to left counts as negative, area under the $x$ axis traversed from left to right counts as negative, and area under the $x$-axis traversed from right to left counts as negative. It's a bit like money: there's no such thing as "negative money", but you can count money as positive when you get it and negative when you owe it.2011-11-24

4 Answers 4

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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 $\int_1^3 f(x) dx=-3$.