calculus
More specifically, if you are given two functions and you are asked to find the area between the two curves is there a way to do it without having to graph them. So far in calculus class we are taught to graph them to see which one is the upper function and which one is the lower function. For example,
If you are given the following two functions is there a way to find which one is the upper function and lower function without having to graph them?
$x=y-\frac{y^2}{4}$ and $x=\frac{y}{4}$
(1) When you integrate with respect to $x$, you’re chopping the region into vertical slices. Look at the shaded region in your picture. First of all, $x$ ranges from $-5$ to $4$, so if you could do the calculation as a single integral, it would be $\int_{-5}^4\text{ something }dx$. But when $x$ is between $-5$ and $0$, vertical slices from from $y=x+2$ at the top down to $y=-\sqrt{x-4}$ at the bottom, while for $x$ between $0$ and $4$ they run from $\sqrt{x-4}$ at the top down to $-\sqrt{x-4}$ at the bottom. In other words, for $-5\le x\le 0$ the slice at $x$ has length
$$\begin{align*}\text{top}-\text{bottom}&=(x+2)-(-\sqrt{x-4})\\ &=x+2+\sqrt{x-4}\;, \end{align*}$$
but for $0\le x\le 4$ it has length $$\begin{align*}\text{top}-\text{bottom}&=\sqrt{x-4}-(-\sqrt{x-4})\\ &=2\sqrt{x-4}\;. \end{align*}$$
Since these are not the same, you have to break the calculation into two parts: the area is
$$\int_{-5}^0(x+2+\sqrt{4-x})\,dx+\int_0^4 2\sqrt{4-x}\,dx\;.$$
Notice that I can’t simply say that $y=\sqrt{4-x}$: $y$ is $\sqrt{4-x}$ for points on the upper branch of the parabola, but for points on the lower branch $y=-\sqrt{4-x}$.
(2) Again, look at the shaded region: when you cut a horizontal slice across it, the slice starts on the left at a point on the straight line and ends at a point on the right on the parabola. For a given $y$, the $x$-coordinate on the parabola is given by $x=4-y^2$, so that’s the $x$-coordinate at the righthand end of the slice; the $x$-coordinate on the straight line is given by $x=y-2$, so that’s the $x$-coordinate at the lefthand end of the slice. The length of the slice is therefore $$\text{right}-\text{left}=(4-y^2)-(y-2)=6-y-y^2\;,$$ and the infinitesimal bit of area contributed by it is its length times its width $dy$:
$$dA=(6-y-y^2)\,dy\;.$$ (For this one you do have the correct limits of integration.)
You'll want to find the point of intersection, and use the $t=x$-value of this point as a bound for integrating (see note below).
Set the two equations equal to one another, and solve for the values $t$ (the x-coordinate) that solves the resulting equation: there will be one such value on your domain.
$$9\sin(t) = 10\cos(t)\implies \frac{\sin t}{\cos t} = \frac {10}{9}\implies \tan t = \frac{10}{9} \implies t = \tan^{-1} \frac{10}{9}$$
Then calculate the sum of the integrals $$\int_0^t (10\cos x - 9\sin x)\,dx + \int_t^\pi (9\sin x- 10\cos x)\,dx.$$
Note: you need to know the point of intersection to divide the integral into two parts, because from $0\leq x \lt t$, where t is the value of x at the point of intersection, $10\cos x > 9\sin x$. At $x = t$, $10\cos x = 9 \sin x$. And from $t \lt x \leq \pi$, $9\sin x > 10 \cos x$. So to measure total area, we need to ensure we choose the correct function as the "upper bound" of each region when we integrate to obtain the total area of the regions bound by the functions.
Best Answer
There is not, for polynomials at least, any need to even see which is on top on the various intervals determined.
Example:$$f(x)=2x^3-6x^2+13x, \\ g(x)=x^3+x^2+x.$$ Begin by factoring the difference: $f(x)-g(x)=x(x-3)(x-4).$ [really not so easy for arbitrary polynomials, but likely works if doing a typical calc exercise.]
We see the zeros of the difference are $0,\ 3,\ 4.$ Since dealing with an area between graphs, the regions before $0$ and after $4$ are not considered (would contribute infinite areas). In the usual method one would see which is on top on each of the intervals $[0,3]$ and $[3,4]$ etc. But since area over each is the integral of the absolute value of $f-g$, which over each is either $f-g$ or $g-f$ [since the crossing point at $3$ was removed], we can complete the process of getting the total area as follows.
Step (1) Get (an) antiderivative $F(x)$ of the difference. Here $F(x)=(1/4)x^4-(7/3)x^3+6x^2.$
Step (2) evaluate $F(x)$ at each intersection point. Here $F(0)=0,\ F(3)=45/4,\ F(4)=32/3.$
Step (3) Add the absolute values of the successive differences. $(45/4-0)+(45/4-32/3)=45/4+7/12=71/6.$
So the area is $71/6,$ and we didn't need to know which function is on top during the intervals between intersections. I think this method also works for other reasonable functions defined on the reals, provided we can get effectively a list of values where the functions are equal, and we can successfully get a closed form antiderivative, and we are not interested in evaluationg improper integrals (which might converge in some cases).