# Why do we subtract things to find the area using the definite integral?

I have already asked a similar question. But the answer in that question is very difficult to understand. I am new to this concept so I am looking for an easier explanation.

My main question is: why do we subtract things to find the area using the definite integral?

Here are a couple of figures -

Area $\displaystyle = \int \left(\sqrt{x} - x^2 \right) dx$

Why do we subtract to find the area? Why not add?

1. Similarly in parabola and line. Area $\displaystyle = \int (x + 2 - x^2)dx$

• It entirely depends on context. Suffice to say, in the examples you give, one would commonly find the area below the blue line over the interval and subtract the area below the red line over the interval in order to "cancel out" the unwanted area which is not shaded in the picture. There are some examples where you would absolutely add different pieces together. – JMoravitz Dec 30 '16 at 7:41
• $\int_a^b f(x)dx$ gives the area under curve $f(x)$ between $x=a$ and $x=b$ above X-axis. – Nitin Uniyal Dec 30 '16 at 7:44
• An explanation like @LoganLuther's above together with a couple of images showing the shapes whose areas are in the subtractions in question, could suffice as an answer. Edit: And maybe the linearity of integration, $\int_a^b f(x) dx - \int_a^b g(x) dx = \int_a^b ( f(x)-g(x) ) dx$ should be mentioned. – Jeppe Stig Nielsen Dec 30 '16 at 7:55
• @JohnSr The tag "conic sections" is not suitable. Please remove. – Logan Luther Dec 30 '16 at 8:06
• What would you like to add? – Jasper Dec 30 '16 at 11:29

## 8 Answers

The given curves are :

y = f(x). .........(1)

y = g(x). ...........(2)

between a and b. So (1) and (2) intersect at x = a and x = b. A thin vertical strip of width dx is taken between the lines x = a and x = b as shown in the figure.

The height of this vertical strip is given as f(x) - g(x) as f(x) $\ge$ g(x) as shown.

So, the elementary area of this strip dA = [f(x) - g(x)] dx.

Now, we know that the total area is made up of vary large number of such strips, starting from x = a to x = b.

Hence, the total enclosed area A, between the curves is given by adding the area of all such strips between a and b:

A = $\int_a^b$ [f(x) – g(x)] dx

The enclosed area between two curves can also be calculated in the following manner,

A = (area bounded by the curve y = f(x), x-axis and the lines x = a and x = b) – (area bounded by the curve y = g(x), x-axis and the lines x = a and x = b)

Hope its clear to you now.

$\int_a^b f(x)$ gives the area "under the curve" of $f(x)$ between $a$ and $b$: the area from the $x$-axis to the curve, across that interval.

In the cases given, you have two curves that you're dealing with instead; one (which I'll call $f(x)$) is higher and the other ($g(x)$) is lower. Finding the area between these curves means finding the area that is under $f$ but not under $g$. This, it is simple to see, we can do by subtraction: the vertical space between $f(x) = \sqrt{x}$ and $g(x) = x^2$ at, say, $1/4$ is equal to $$f\left(\frac{1}{4}\right)-g\left(\frac{1}{4}\right) = \frac{1}{2} - \frac{1}{16} = \frac{7}{16}$$ because while there's $1/2$ below $f$ there, $1/16$ is also below $g$ and thus shouldn't get counted.

Here's this pair of functions with their integrals displayed, overlapping. The vertical space mentioned above is also shown as a vertical line. The area under $f$ contains both the pale area and the darker area; the area under $g$ contains only the darker area. But we need only the pale area; we can find both (by taking the integral of $f$), and then remove the dark area (by taking the integral of $g$ and subtracting that from the integral of $f$).

• I am not clear with vertical space. Where it is? – user404716 Dec 30 '16 at 12:15
• The drawing now includes the vertical line that matches the vertical space. – Dan Uznanski Dec 31 '16 at 11:56

Look at the second image that you provided. Let $A$ represent the region under the line; let $B$ represent the region under the parabola; and let $C$ represent the region in between the line and the parabola. Note that $B$ and $C$ do not overlap. We want to find the area of region $C$.

Notice that the combination of regions $B$ and $C$ completely covers region $A$. Restated: $$B\cup C=A\tag{1}$$ Equation $(1)$ means that all points in $A$ lie in $B$ or in $C$. We can solve for the region $C$. $$C=A-B\tag{2}$$ Equation $(2)$ means that all points in $C$ lie in $A$ but do not lie in $B$.

The area of a region is simply the sum of all the points in that region.

• Also you must state that $B\cap C = \emptyset$, i.e, $B$ and $C$ are disjoint. – Max Payne Dec 30 '16 at 8:08
• @MaxPayne Thanks for that addition. – pseudoeuclidean Dec 30 '16 at 8:14

In this context, to find the area shaded, think about the area under each curve separately, and then try to find the area NOT shared by them. What will you get?

Since the area is a plain number, to find a difference you simply subtract two areas: You can even think of it in terms of apples. Suppose you have $15$ apples, arranged in a $3×5$ grid.

Now for any set of apples, the total number of apples is not about how they are arrranged, and therefore for any number of apples that we remove from the grid, we do not need to care about the way they were in the grid, but only the number removed.

And to see also why:

$$\int_a^b (f+g)(x) \mathrm dx = \int_a^b f(x) \mathrm dx + \int_a^b g(x) \mathrm dx.$$

Observe that the the definite integral is a linear operator, meaning it distributes over addition and this can be verified using the fundamental theorem of calculus.

$$h(x) =(f+g)(x)=f(x)+g(x),$$

$$\int_a^b h(x) \mathrm dx =H(b)-H(a)=$$ $$F(b)+G(b)-G(a)-F(a).$$

Rearranging terms gives:

$$F(b)-F(a)+G(b)-G(a)=\int_a^b f(x) \mathrm dx +\int_a^b g(x) \mathrm dx.$$

If you decompose the area in thin vertical rectangles, the height of the rectangles is the difference between the ordinates. Come on! Some of us use integrals, or even the fundamental idiom of integral calculus to explain the idea that areas can be subtracted. However, the concept of subtracting areas is much more general, and much older, than this. An area is an area; integrals are just a means to calculate them; more often they aren't even used that way. If I'm not mistaken, subtraction of areas goes all the way back to Euclid.

As an example. Starting with a rectangle (with even more rectangles in it), we can do the following to deduce the area of an arbitrary triangle. Herewith it is only assumed that the formula for the area of a rectangle is well understood.
To begin with, a parallelogram $\overline{ABDC}$ is constructed inside a rectangle $\overline{AEDF}$: Label the vertices and assign a few coordinates (addition of vectors giving $D$): $$A = (0,0) \quad , \quad B = (x_1,y_1) \quad , \quad C = (x_2,y_2) \quad \Longrightarrow \quad D = (x_1+x_2,y_1+y_2)$$ Then calculate areas according to the following, leaving it to the user to prove things where they feel the need. Though, with Euclidean geometry, What you see is what you get, most of the time: $$\operatorname{Area}(\overline{CDf}) = \operatorname{Area}(\overline{AbB}) = \frac{1}{2} \operatorname{Area}(\overline{AbBh}) = \frac{1}{2} \overline{Ab}\times\overline{Ah} = \frac{1}{2} x_1y_1 \\ \operatorname{Area}(\overline{BcD}) = \operatorname{Area}(\overline{ACg}) = \frac{1}{2} \operatorname{Area}(\overline{AaCg}) = \frac{1}{2} \overline{Aa}\times\overline{Ag} = \frac{1}{2} x_2y_2 \\ \operatorname{Area}(\overline{bEcB}) = \operatorname{Area}(\overline{gCfF}) = \operatorname{Area}(\overline{Aakh}) = \overline{Aa}\times\overline{Ah} = x_2y_1$$ Now subtract the areas of $\overline{AbB} , \overline{bEcB} , \overline{BcD} , \overline{CDf} , \overline{gCfF} , \overline{ACg}$ from the area of the big rectangle, which is $\operatorname{Area}(\overline{AEDF}) = (x_1+x_2)(y_1+y_2)$. Then what we get is the area of the parallelogram: $$\operatorname{Area}(\overline{ABDC}) = (x_1+x_2)(y_1+y_2) - 2 x_2y_1 - 2\cdot\frac{1}{2} x_2y_2 - 2\cdot\frac{1}{2} x_1y_1 =\\ x_1y_1+x_1y_2+x_2y_1+x_2y_2 - x_2y_2 - x_1y_1 - 2x_2y_1 = x_1y_2-x_2y_1$$ Do we observe a determinant here? Ayway, the area of the triangle is half of this: $$\operatorname{Area}(\overline{ABC}) = \frac{1}{2}(x_1y_2-x_2y_1) = \frac{1}{2} \begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix}$$ There are many interesting applications. For example, the area of an arbitrary polygon can be calculated with addition and subtraction of nothing else but triangles, as exemplified here.

However, there is a tag (definite-integrals) associated with the question, so let's kill the mosquito with a gun end employ some of that in the end: $$\operatorname{Area}(\overline{ABC}) = \operatorname{Area}(\overline{AaC}) + \operatorname{Area}(\overline{abBC}) - \operatorname{Area}(\overline{AbB}) =\\ \int_0^{x_2} \frac{y_2}{x_2}x\,dx \; + \; \int_{x_2}^{x_1} \left[y_2 + (y_1-y_2)\frac{x-x_2}{x_1-x_2} \right] dx \; - \; \int_0^{x_1} \frac{y_1}{x_1}x\,dx = \\ \frac{1}{2}x_2y_2 + \frac{1}{2}(x_1-x_2)(y_2+y_1)-\frac{1}{2}x_1y_1 = \frac{1}{2}(x_1y_2-x_2y_1)$$ Always nice to see the consistency of mathematics showing up again.

• Just a minor point, but I recommend using "subtract" instead of "substract" (see english.stackexchange.com/questions/3640/… for some background). – J W Jan 3 '17 at 14:44
• @JW: You're welcome. Think I've corrected these. Don't hesitate to point out other mistakes if there are some: English is not my mother's tongue. Maarruh .. wat schuift het? :-) – Han de Bruijn Jan 3 '17 at 18:27
• English please :( – Max Payne Jan 4 '17 at 14:24

It is not complicated. One common area is falling on a similar area as overlap, some addition is involved and where from the minus sign? that seems to be your doubt ... which you expressed nicely with a graphic.I edited your picture to explain as follows. We have Area = height H multiplied by width for a rectangle standing on x-axis. But the height is seen variable now as $(y_2-y_1)$. In the sideways bulged patch this height H goes from zero to a maximum and then back to zero. and its constant width is $dx$. Area is summation by integration..

Suppose the topmost graph is 1 then 2, 3, 4 . We have to find the area between the Red-Curve and Blue line in the first quadrant ( Graph 1). When we Integrate the Red-Curve with the limit from 0 to some positive x , We get the area of red region ( Graph 2). In the same manner, When we Integrate blue line with limit from 0 to some positive x we get the area of blue region ( Graph 3). To find the area between Blue-Line and Red-Curve, We Integrate the Blue-Line with the limit x = 0 to x = 2. Similary we Integrate Red-Curve with limit x = 0 to x = 2 . Since Blue-Line is above the Red-Curve, Subtract the Blue Area from the Red Area To get Green Area ( Graph 4 )

• You can attach link. – user404716 Jan 1 '17 at 17:01
• @JohnSr Read the text. You will get it. I joined this site today only therefore, I am learning how to do stuffs – user39390 Jan 1 '17 at 17:07
• You have option to add pics. So add it. – user404716 Jan 1 '17 at 17:43
• @JohnSr That need 10 reputations and I don't have. – user39390 Jan 1 '17 at 17:43
• @JohnSr Yeah Now look at the answer. – user39390 Jan 1 '17 at 17:51