Average Values

Think about the definition of average velocity: $v=\frac{\Delta s}{\Delta t}$, or, the change in displacement over the change in time. Since displacement is the integral of velocity, we can deduce the first application of the integral: the average value of a function. This is actually really straightforward.

The average value of any function $f(x)$ is equal to

$f(x)_{avg}=\displaystyle\frac{\int_a^bf(x)dx}{b-a}$

This gives rise to an interesting theorem: the Mean Value Theorem for integrals. This states that on the interval $a$ to $b$, for a continuous function there will be a point where the value of the function will be equal to the average value of that function from $a$ to $b$. This can be represented graphically by drawing a secant line between $f(a)$ and $f(b)$ (which represents the function's average value on that interval) and finding the point where the tangent line is parallel to it.

Area Between Curves

We know that the integral is used to find the area under the curve. In reality, since the area can be negative, its the area between the curve and the $x$-axis. But what if we wanted to find the area between a curve and another curve? If we think of this graphically, the answer seems obvious.

If we wanted to find the shaded area, all we have to do is subtract the integrals of the functions! Thus one formula for the area between two curves is $\displaystyle\int_a^b[f(x)-g(x)]dx$

It should be noted that it is necessary to subtract the "lower" function from the "higher" function. If you subtract backwards, you will get negative the correct answer.

Since we are rarely get graphs, finding which function is the higher can often be difficult. Thus, we can take a more rigorous approach which involves Riemann sums and has several benefits.

Imagine the same area between two curves, but this time it is divided into many thin strips. As the width of the strips gets smaller and smaller, the approximation for the area gets better and the strips go to rectangles. This is a Riemann sum. Thus, we can represent the total area, $A$, between the curves in terms of the areas of the individual rectangles, which we will call $dA$.

In terms of our functions, $f(x)$ and $g(x)$, the area of an individual rectangle, $dA$ is $[f(x)-g(x)]dx$ where $f(x)-g(x)$ is the height of the rectangle, and $dx$ is the change in $x$, or the width of the rectangle. If we sum up all the areas, we arrive with $\displaystyle\lim_{n \to \infty}[f(x)-g(x)]dx$, which is really the same thing as $\displaystyle\int_a^b[f(x)-g(x)]dx$ as we saw earlier. However, instead of taking the integral directly, we just used another technique called **slicing** to arrive at the same result. There are benefits to this method, as it provides a bridge to more complicated functions.

Suppose we had a curve that had more than one output for one input. If we sliced the function vertically, then we would be finding the integral between the same curve twice. Though this can be done, it is much easier to employ another technique: slicing horizontally.

In this case, the "height" of the rectangles is a function of $y$ and the "width" is $dy$. Notice also that the bounds of integration are along the $y$-axis for horizontal slices.

So, the formula becomes $\displaystyle\int_{y_a}^{y_b}[f(y)-g(y)]dy$. Remember, to get from a function of $x$ to a function of $y$, all we have to do is take the inverse: if $y=f(x)$ then $x=f^{-1}(y)$.

Rotated Solids

Some visual imagination is required for this, but we soon realize that we can make a 3-dimensional object by rotating a function about some line. Since we have a solid when we do this, we can do fun stuff like find its volume using integrals.

Disk Method: Now we can see what all the fuss with slicing was about: finding the volume isn't as simple as subtracting two integrals. Instead, you must find the volume using strips with volume $dV$. Consider the function $y=e^x$ rotated about the $x$-axis. This will form a trumpet-like shape. If a function is rotated about the $x$- axis, then it is most convenient to do a vertical slice. When we do this, we see that we get thin, ring-like pieces making up the solid. As the number of slices approaches infinity, these rings (technically called ''frustums") will tend to the shape of a disk (aka cylinder).
Remembering some geometry, we recall that the volume of a cylinder is $V=\pi r^2h$. Thus, we can write the volume of $dV$ as $dV=\pi y^2dx$, where the function $y$ at any point on the curve is the radius, and $dx$ is the height.

Now using the same approach as with areas to integrate, we can find the volume of the whole figure by simply integrating $dV$ to get the whole volume $V$ (without proof, $dV$ is the derivative of $V$): $V=\pi\displaystyle\int_a^by^2dx$. This will always be the case for solids which are rotated uniformly about some line, as they will always form cylindrical cross-sections.

Again, as with areas, we can find the volume of a solid rotated about the $y$-axis by simply realizing that the radius of the cylinder is $f(y)^2$ and the height is $dy$. Thus, the integral becomes $V=\pi\displaystyle\int_{y_a}^{y_b}x^2dy$. (Note that $f(y)=x$). This applies to any sort of solids rotated about a vertical asymptote.

Washer Method: Now imagine if you were presented with two intersecting functions that were both being rotated about a common point. In this scenario, you would be asked to find the volume between the two. Examining a diagram, we can see that the cross-sections of this figure will look like disks with varying sizes of holes in them, otherwise known as "washers".

The key insight here is that the volume of the cross section, $dV$, can be obtained by subtracting the function with the smaller radius($r$) from that of the larger radius($R$): $dV=\pi R^2dx-\pi r^2dx$, which can be simplified into the generally-used form $dV=\pi(R^2-r^2)dx$. Using the same as with regular rotated solids, we can now integrate $dV$ in order to get $V$: $V=\displaystyle\int_a^bdV=\displaystyle\int_a^b\pi(R^2-r^2)dx$. This formula is used for all washers.

Slabs: Finally, we can have a solid that doesn't have cylindrical cross-sections. This doesn't often come up, as it isn't a "rotated" solid, but it is an application of the integral nonetheless. You should realize that the volume of any right solid (a cross-section will always be "right", which just means that its "height" edges are perpendicular to its base) is the area of its base multiplied by its height: $V=A_{base}\times h$. This way, it doesn't matter what sort of solid you have, be it a pyramid or a dodecagonal prism, all you have to do to find its volume is find the volume of its cross-section (in terms of the function that defines it, as we have already been doing), and integrate it to get the total volume.