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See this link to get a picture of what I mean.

If you want to calculate the area between $f(x)$ and $g(x)$ on a certain interval $[a,b]$. Do I have to add the two areas between the $x$-axis and $f(x)$, the $x$-axis and $g(x)$, or can I just find the area between $f(x)$ and $g(x)$ in one step like this:

$$\int_a^b\left[f\left(x\right)-g\left(x\right)\right]dx$$

My question is, how do you prove the above will work, since $g(x)$ has negative values but $f(x)$ has positive values. Will this be giving you the incorrect area between $f(x)$ and $g(x)$?

Sahiba Arora
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    $g(x)$ is negative on the given interval, therefore you are subtracting a negative function, which in result will give you a positive area. Think of $\int_a^b g(x),dx=A_1$ where $A_1<0$ and $-\int_a^b g(x)=-A_1$ where now the area is positive. In the integral that you wrote you are summing the two positive areas formed by $f(x),g(x)$ and the $x$ axis. – kiyomi Feb 02 '18 at 21:21

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Absolutely not, this does not give the incorrect area. To prove it, we have to translate both $f(x)$ and $g(x)$ up $y$ units, so that both $f(x)$ and $g(x)$ are positive. This will be represented like this:

$$\int_a^b\left[(f\left(x\right)+y)-(g\left(x\right)+y)\right]dx$$

Here, you are not changing the area since you are only translating the graph $y$ units towards the positive $y$ axis.

The expression above would simplify to:

$$\int_a^b\left[f\left(x\right)-g\left(x\right)\right]dx$$

Therefore, the area enclosed by $f(x)$ and $g(x)$, where $f(x)>g(x)$ is defined as:

$$\int_a^b\left[f\left(x\right)-g\left(x\right)\right]dx$$

Yash Jain
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