areacalculusintegration
Find the area enclosed by the curves $y = \tan(x)$, $y = -3\tan(x)$ and $x = \pi/4$.
What can possibly be the area, $\tan(x)$ and $-3\tan(x)$ never intersect unless it is $\frac{\pi}{2}k$. There is no upper bound to find because the two curve goes to different direction, so confused about this question.
From Wolfram Alpha, we can sketch the curves to find the area of interest:
Note that we need to find the points of intersection: at $x = 0$ the lines $y = x, \;y = \frac x{4}$ intersect. At $x= 1$, the lines $y = x$ and $y = \frac 1x$ intersect. At $x = 2,$ the lines $y = \frac 1x $ and $y = \frac x4$ intersect. You can solve this by integrating between the relevant curves from $x = 0$ to $x = 1$, and likewise integrating between the relevant curves between $x = 1$ and $x = 2$, then summing: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx $$
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.
Best Answer
HINT
The integral you are interested in can be written as \begin{align*} I = \int_{0}^{\pi/4}(\tan(x) - (-3\tan(x)))\mathrm{d}x = \int_{0}^{\pi/4}4\tan(x)\mathrm{d}x \end{align*}
Can you take it from here?