Find the area between $y=x$ and $y=x^2$.

Solution

Since no bounds were given, the only interpretation is that we're looking for the area enclosed within the two curves between the two points where they intersect.

Notice that between these intersection points, the upper function is $f(x)=x$ and the lower function is $g(x)=x^2$, so to calculate the area, we'll find the difference $f(x)-g(x) = x-x^2.$

We need to find where the two functions intersect; to do so, set them equal to each other and solve: $\begin{aligned} x &= x^2\ 0 &= x^2-x\ 0 &= x(x-1)\ x &= 0, 1 \end{aligned}$

Notice how we solved this like any other quadratic equation, by moving all the terms to one side and factoring. Many students will look at the first line and instinctively want to divide both sides by $x.$ This gives one of the answers ($x=1$), but it misses the second ($x=0$). Why is this? Remember that division by zero is meaningless, so when you divide both sides of an equation by $x$, you're assuming that $x \neq 0$. Thus, doing so will give you all the answers where $x$ is not $0$, but you may miss the fact that $x=0$ is also a solution.

Thus the area between the two curves is $\begin{aligned} A &= \int_0^1 x-x^2\ dx\ &= \dfrac{1}{2}x^2 - \dfrac{1}{3}x^3 \bigg|_0^1\ &= \dfrac{1}{2}-\dfrac{1}{3}\ &= \ans{\dfrac{1}{6}} \end{aligned}$

Using Horizontal Slices

Find the area between $y=x$ and $y=x^2$ using horizontal slices.

Solution

Notice what changes: since the slices are horizontal, the thickness of each rectangle will be $\Delta y$ instead of $\Delta x$. Thus in the final integral, we'll have $dy$ as the differential instead of $dx$.

That means that we need the functions written in terms of $y$ instead of $x$. The first function is easy; we don't actually have to do any work, since $y=x$ is the same as $x=y$. For the second function, solve for $x$: $y=x^2 \longrightarrow x = \pm \sqrt{y}$ Since we only need the positive part, we can drop the plus/minus.

This leaves two questions:

In what order do we subtract the two functions?

Instead of upper and lower functions, we have right and left functions, but the principle is the same: take the larger (in this case the right) and subtract the smaller (the left): $\sqrt{y} - y$

What are the limits of integration?

The limits of integration, like the functions and differential, will be in terms of $y$: the lowest $y$ value is $0$, and the highest $y$ value is $1$. Thus the area is $\begin{aligned} A &= \int_0^1 \sqrt{y}-y \ dy\ &= \dfrac{2}{3}y^{3/2} - \dfrac{1}{2}y^2 \bigg|_0^1\ &= \dfrac{2}{3} - \dfrac{1}{2}\ &= \ans{\dfrac{1}{6}} \end{aligned}$

Find the area between $x=2y-y^2$ and $x=y^2-4y$.

Solution

In all of these problems, we have the same two questions: what's the order of subtraction, and what are the limits of integration?

By looking at the picture above, the "upper" (rightmost) function is $x=2y-y^2$, so the integral will involve $2y-y^2 - (y^2-4y) = -2y^2 + 6y$

To find the limits of integration, set the two functions equal to each other and solve: $\begin{aligned} 2y-y^2 &= y^2-4y\ -2y^2+6y &= 0\ 2y(-y+3) &= 0\ y &= 0, 3 \end{aligned}$

Thus, the area is $\begin{aligned} A &= \int_0^3 -2y^2 + 6y \ dy\ &= -\dfrac{2}{3}y^3 + 3y^2 \bigg|_0^3\ &= -18+27\ &= \ans{9} \end{aligned}$

Find the area between $y=x^3$ and $y=x$ on the interval $[-1,2]$.

Solution

There are a couple of new things in this example: we're given explicit limits, and also, the curves intersect several times.

Whenever the curves intersect, the upper and lower functions swap. This means that we need to split this area into several integrals (three in this case -- from $-1$ to $0$, from $0$ to $1$, and from $1$ to $2$).

The total area is the sum of these three integrals: $\begin{aligned} A &= \int_{-1}^0 x^3-x \ dx + \int_0^1 x-x^3\ dx + \int_1^2 x^3-x\ dx\ &= \left[\dfrac{1}{4}x^4-\dfrac{1}{2}x^2\right]_{-1}^0 + \left[\dfrac{1}{2}x^2-\dfrac{1}{4}x^4\right]_0^1 + \left[\dfrac{1}{4}x^4-\dfrac{1}{2}x^2\right]_1^2\ &= \left[0-\left(\dfrac{1}{4} - \dfrac{1}{2}\right)\right] + \left[\dfrac{1}{2}-\dfrac{1}{4}\right] + \left[\left(\dfrac{16}{4}-\dfrac{4}{2}\right) - \left(\dfrac{1}{4}-\dfrac{1}{2}\right)\right]\ &= \dfrac{1}{4} + \dfrac{1}{4} + 2 + \dfrac{1}{4}\ &= \ans{\dfrac{11}{4}} \end{aligned}$