Find the area enclosed by $y=x$, $y=3x$ and $y=4-x$.
Looking at the graph, $$A = \int_0^2 (3x)-(4-x)\, dx – \int_{0}^2 x\, dx=-2$$
Clearly the area is above the $x$-axis, but why $A<0$?
[Math] Find the area enclosed by the lines
areacalculusdefinite integralsintegration
Related Solutions
It is a reasonable question. Although usually if asked to find the area of a region or regions bounded by two graphs what is meant by "bounded" is that the regions all lie within the interior of some circle.
This is analogous to a bounded set on the number line being contained in some interval $[a,b]$. It is completely circumscribed.
However it is possible for to graphs to enclose a finite, yet unbounded region.
There are many examples, but one is as follows.
Find the area of the region "bounded" by the graphs of $y=0$ and $y=\dfrac{x}{x^4+1}$
Here is the graph of the region.
This region is not bounded in the sense stated above. It cannot be contained in the interior of a circle. Yet it has a finite area.
\begin{equation} \int_{-\infty}^\infty\dfrac{|x|}{x^4+1}\,dx=\int_{0}^\infty\dfrac{2x}{x^4+1}\,dx\\ \end{equation}
Make the substitution $u=x^2$, $du=2x\,dx$ and this becomes
\begin{eqnarray} \int_{0}^\infty\dfrac{1}{u^2+1}\,du&=&\frac{1}{2}\arctan(u){\Large\vert}_{0}^\infty\\ &=&\left(\dfrac{\pi}{2}-0\right)\\ &=&\frac{\pi}{2} \end{eqnarray}
Therefore it is acceptable to say that, in a sense, an unbounded region is "bounded" by two graphs so long as the area enclosed is finite.
By symmetry, it is twice the area between the $y$-axis and the line $x=2$. The positive $x$-intercept is $x=1$, and on the interval $[0,1]$, the curve is below the $x$-axis, hence the total (geometric) area is $$2\biggl(\int_1^2y(x)\,\mathrm d x-\int_0^1y(x)\,\mathrm d x\biggl).$$
Best Answer
The limits of integration are incorrect. The integral should be $$\begin{align} A&=\int_0^1 3x dx + \int_1^2 (4-x) dx - \int_0^2 x dx \\ &=\frac{3}{2} + 8 - 2 - 4 + \frac{1}{2} -2 \\ &=2 \end{align}$$