# Area of a region bounded between two curves

Please refer to the following image. Based on the formula, graphically speaking, to obtain the area of the region bounded between the two curves $y=f(x)$ and $y=g(x)$, I am supposed to subtract the area of the shaded region in graph $2$ from the area of the shaded region in graph $1$ so that I can get the area between the two curves.

I can understand why I should subtract that small portion (bounded by the graph $y=g(x), x=b$, and the $x$-axis), but I cannot understand subtracting the portion (which lies below the x-axis) bounded by $y=g(x)$ and $x=a$ and $x=c$, where c is the $x$-coordinate of the point where the graph $y=g(x)$ cuts the $x$-axis. I cannot understand why this must be subtracted because this portion was never part of the shaded region of $y=f(x)$ anyway! It seems as though I am subtracting something unnecessarily! Can somebody explain this, visually if possible?

Note: In both graphs $1$ and $2$, I have included the other graph in dotted lines.

## 2 Answers

The way the problem is set up you shouldn't subtract the area above $g(x)$ from the area under $f(x)$ geometrically speaking because areas are not defined to be negative(which why we often use absolute values in such situations). In fact to get the area of the shaded region between the two should be added. Remember that the area given by an integral with respect to $x$ is the area between the bounded region and the $x$-axis.

However, the so-called "area" above $g(x)$ will be negative so you will end up adding the two integrals instead of subtracting. The integral of a curve below the $x$-axis is always a negative value so you get the positive value: $$\int^{b}_{a}f(x)dx -(-\int^{b}_{a}g(x)dx)$$ $\implies$ $$\int^{b}_{a}f(x)dx +\int^{b}_{a}g(x)dx$$ Thus, you do not end up subtracting. Another way is to see the two areas in terms of absolute value which if you visualize the following should makes things clear:$$|\int^{b}_{a}f(x)dx| +|\int^{b}_{a}g(x)dx|$$

And, if you must perform/understand the subtraction geometrically then one way to do it would be to add some constant $c$ to both function such that $g'(x) = 0$ (to make sure the minima touches the $x-axis$) for some $x \in [a,b]$ Basically if you add some constant you raise the graphs of the functions up so the your $g(x)$ has it's lowest point touching the $x$-axis giving you set up you need and then subtract the integral value of $g(x)+c$ from the integral value of $f(x)+c$. For example, below $c = 4$ and it has been added to both functions so you as you can probably eye ball the area between them is the same, although in the case of the functions with the added constant you will be subtracting positive areas. This will be subtraction of two positive areas in the conventional sense of geometry. where the blue graph becomes the green graph and the red becomes the black one.

Hope that helps.

• Just some phrasing issues I wish to clarify :) What do you mean by "add some constant c to both function such that g′(x)=0". Do you mean like...add some constant c to both function such that there is a minimum point on the graph g(x) [or whichever the "lower graph" is], and this minimum point touches or is above the x-axis? Because g'(x) can still, at a particular instant, be zero, even if the minima does not touch the x-axis. Furthermore, g'(x) can never always be zero - unless graph of g(x) is a horizontal line - so I think you meant g'(x) at a particular instant? – Charlz97 Dec 12 '16 at 2:21
• Yes your reasoning is correct. And I made a mistake, I could of said that the minima of lower function only be greater or equal to $0$ rather than equal to 0.So the minima doesn't necessarily need to touch the x-axis. And yes I mean for some value(s) of $x$ not all(unless like you said it is a constant function) so that $g'(x)=0$. – Red Dec 12 '16 at 2:55
• Hello can you please tell me the site link where you draw this graph. I will be very thankful to you. – Kanwaljit Singh Dec 12 '16 at 17:06
• @KanwaljitSingh Sure, it is called Desmos graphing calculator desmos.com/calculator This is a link from my phone so in case there is a problem you can easily find it through a google search. Good Luck! – Red Dec 12 '16 at 17:15
• @Kanwaljit Singh Sure, it is called Desmos graphing calculator desmos.com/calculator This is a link from my phone so in case there is a problem you can easily find it through a google search. Good Luck! – Red Dec 12 '16 at 17:17

I think you are right that it should not subtract the area which lies below the x axis and It should add this area. The fact is that the $\int_a^cg(x)dx$ is not positive but a negative value. So "subtracting" ends up summing up these two areas.