calculusintegration
I know this is a very elementary question but I can't make out the answer from the other posts I found in my search.
These are three lines, I need to find the area enclosed by them. how do I go about it? Do I split it into two halves with the intersection point of -x + 10 and y = 4x as the midpoint? is there any simpler way?
Also, what would be the way if these were curves?
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: $$\int_0^1 \left(x - \frac x4\right)\,dx \quad + \quad \int_1^2 \left(\frac 1x - \frac x4\right)\,dx $$
Best Answer
Yes, as you guessed, a good way to do the integral is to find the intersection of the lines $y=-x+10$ and $y=4x$ (this occurs at $x=2$), and divide the area into the portion to the left of the line $x=2$ (bounded by $y=4x$ above and $y=2x$ below) and the portion to the right of the line $x=2$ (bounded by $y=-x+10$ above and $y=2x$ below).
You can then solve this in two integrals.
Another answer shows how to find the area using three integrals, again using the line $x=2$ to divide a single region into two.
The methods are pretty much the same if the bounds are curves rather than straight lines, unless there are additional intersections of the bounds to be concerned about. (Two distinct lines can intersect only once; two distinct curves can intersect multiple times.) But suppose (for example) you had the bound $y=\frac{16}{x}$ instead of $y=-x+10$; you'd still be able to use two integrals, one to the left of $x=2$ and one to the right.
Just for fun, let's consider what happens if we apply the transformation $(x,y) \mapsto (x,y-2x)$ to this figure. This transformation happens to preserve area, so the resulting triangle, bounded by the lines $y=-3x+10$, $y=2x$, and $y=0$, has the same area as the original triangle.
The first two bounds intersect at $(2,4)$. Let's put the equations of those two lines in terms of $y$ and integrate over $y$ from $0$ to $4$:
$$y=-x+10 \implies x = -\frac13(y-10) = \frac{10}{3} - \frac13 y;$$ $$y=2x \implies x = \frac12 y;$$ $$\begin{eqnarray} A &=& \int_0^4 \left(\left(\frac{10}{3} - \frac13 y\right) - \frac12 y\right)\,dy \\ &=& \int_0^4 \left(\frac{10}{3} - \frac56 y\right)\,dy \\ &=& \left.\left(\frac{10}{3}y - \frac{5}{12} y^2\right) \right|_0^4 \\ &=& \frac{20}{3}. \end{eqnarray}$$