I try to figure out the area between the function $f(x)=x^2$ and function $g(x)=x+6$ bounded by lines $x=0$ to $x=6$. I know, how will the area look like, but I've got problem with subtraction or addition of integrals.
This is what I figured out so far:
In bound from 0 to 6 function $f$ and function $g$ have one point in common and that is the point $P=[3,9]$.
In bound from 0 to 3 the function $g$ is above the function $f$, so I got integral:
$\int_{0}^{3}(g-f) dx=\int_{0}^{3}(x+6-x^2)dx=[\frac{x^2}{2}+6x-\frac{x^3}{3}]_0^3 = \frac{27}{2}$
In bound from 3 to 6 the function $f$ is above the function $g$, so I got integral:
$\int_{3}^{6}(f-g) dx=\int_{3}^{6}(x^2-x-6)dx=[\frac{x^3}{3}-\frac{x^2}{2}-6x]_3^6 = \frac{63}{2}$
The result has to be 18, but with my results can I get 18 just by subtracting the second area from the first area.
Why should I subtract and not to sum up the areas?
1 Answer
When they want the "area between the two curves", then you have to add the two numbers you got. This area is the $L^1$-distance of the two functions $y_1(x):=x^2$ and $y_2(x):=x+6$ over the interval $[0,6]$.
When they wanted $18$ they might as well have asked for the integral $\int_0^6\bigl(y_1(x)-y_2(x)\bigr)\>dx$. This integral represents the overall difference in the "effectivity" of the two functions $y_1$ and $y_2$ over the interval $[0,6]$ (say, sales of two firms in the months January–June).
