Subsection 8.2.1 Definition
Recall the distance travelled problem described in
Subsection 6.1.2, where I tried to understand the distance travelled by an object when I knew its velocity function
\(v(t)\text{.}\) Geometrically, this was equivalent to finding the area under the graph of the function.
I set up a process to solve this problem, at least approximately. That process involved making rectangles under the curve and adding up the area of the rectangles, as in
Figure 6.1.8. Integration will be the limit of this approximation process, just as differentiation was the limit of the approximation process of secant lines approaching the tangent line.
Sigma notation is a concice notation for the approximation of areas by rectangles. I am trying to calculate the area under the curve of \(f(x)\text{,}\) a positive function, defined on an interval \([a,b]\text{.}\) I divide the area into \(n\) rectangles; if I divide equally, the width of each rectangle will be \(\frac{b-a}{n}\text{.}\) The height of the rectangle will be the function value \(f(x^*)\) where \(x^*\) is some particular \(x\) in the subinterval.
With these notations, the area of a rectangle is width times height, that is \(\frac{b-a}{n} f(x^*)\text{.}\) To be more specific, if I use \(k\) as an index to refer to various rectangles and \(x_k^*\) is in the kth interval, the area of the \(k\)th rectangle is \(\frac{b-a}{n} f(x_k^*)\text{.}\) To get the complete approximation, I add up all these rectangles with sigma notation.
\begin{equation*}
A \cong \sum_{k=1}^n \frac{b-a}{n} f(x_k^*)
\end{equation*}
This sum is called a Riemann sum approximation for area. If the function is well behaved (continuous guaranteees this), then I can keep taking finer and finer partitions for better and better approximations. This process can be extended into some kind of limit process. In the limit, I hope for a perfect calculation of area.
\begin{equation*}
A = \lim_{n \rightarrow \infty} \sum_{k=1}^n f(x_k^*)
\frac{b-a}{n}
\end{equation*}
This limit, if it exists, is called a definite integal. The area under the curve is calculated by a definite integral. Since this definition uses Riemann sums, this is called the Riemann definiton of the integral, or simply the Riemann integral. It has a new notation.
\begin{equation*}
\int_a^b f(x) dx := \lim_{n \rightarrow \infty} \sum_{k=1}^n
f(x_k^*) \frac{b-a}{n}
\end{equation*}
I should explain the notation. The sigma of sigma notation gets replaced with
\(\int\text{,}\) which is the integral sign. The bounds
\(k=1\) and
\(n\) get replaced with the endpoints of the interval,
\(a\) and
\(b\) respectively. The
\(f(x^*_k)\) just becomes the function
\(f(x)\text{.}\) Finally, the width
\(\frac{b-a}{n}\) gets infinitesimally small as
\(n
\rightarrow \infty\text{,}\) so it becomes the infinitesimal term
\(dx\text{.}\) Like Leibniz notation, this
\(dx\) is a relic from a different historical presentation of calculus. Its main purpose in the modern notation is to indicate what symbol is the variable. However, as will be seen in
Section 9.1, this
\(dx\) notation is useful for a variety of integration calculations.
The first problem with limit definition is the existence of the limit. Unfortunately, this is an extremely difficult question to solve in general, particularly since it has to work for any possible choices of the values \(x_k^*\text{.}\) Fortunately, this limit will always exist if \(f(x)\) is continuous, a fact that is presented here without proof.
I should point out some important observations. First, this integral returns a number: given a function and endpoints, it calculates a fixed area. It isn’t (yet) a new function. Second, I need the bounds: the Riemann integral doesn’t make any sense without the bounds of the interval \([a,b]\text{.}\) Third, I defined this for positive functions. It also works for functions which have negative values, but areas below the \(x\) axis are interpreted as negative. If a function has both positive and negative values, the integral will give the difference between the area above the axis (the positive area) and the area below the (the negative area).
Recall when I talked about the derivatives, I mentioned that the definition was correct and rigorous but difficult to use. That comment is also true here, but even more severe. The definition of the integral is nearly impossible to use for most functions. I will only try to use the definition for small order polynomials.
Example 8.2.1.
\begin{align*}
\int_0^3 x^2 dx \amp = \lim_{n \rightarrow \infty}
\sum_{k=1}^n (x^*_k)^2 \frac{b-a}{n}
\end{align*}
I start with the setup of the Riemann integral. I need to make a choice for \(x_k^*\text{.}\) I’ll take the standard choice of \(x_k^* = a + k \frac{b-a}{n}\text{,}\) which in this case is \(x_k^* = 0 + k \frac{3}{n} = \frac{3k}{n}\text{.}\)
\begin{align*}
\amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left(
\frac{3k}{n} \right)^2 \frac{3}{n}
\end{align*}
Then I apply the square to each piece of the term in brackets. Anything that doesn’t involve \(k\) is independent of the sum and can be factored out.
\begin{align*}
\amp = \lim_{n \rightarrow \infty} \frac{27}{n^3}
\sum_{k=1}^n k^2
\end{align*}
I evaluate the sum, which is one of the standard sums.
\begin{align*}
\amp = \lim_{n \rightarrow \infty} \frac{27}{n^3}
\frac{n(n+1)(2n+1)}{6}
\end{align*}
I simplify the expression by expanding the multiplication. Then I can evaluate this as an asymptotic analysis limit. The asymptotic orders are the same, so the limit is the ratio of the leading terms. This will complete the integral.
\begin{align*}
\amp = \lim_{n \rightarrow \infty} \frac{9}{2}
\frac{2n^3+3n^2+n}{n^3} = 9
\end{align*}
Example 8.2.2.
\begin{align*}
\int_2^5 (x^3-x) dx \amp = \lim_{n \rightarrow \infty}
\sum_{k=1}^n \left( (x^*_k)^3 - x^*_k \right)
\frac{b-a}{n}
\end{align*}
I start with the setup of the Riemann integral. I need to make a choice for \(x_k^*\text{.}\) I’ll take the standard choice of \(x_k^* = a + k \frac{b-a}{n}\text{,}\) which in this case is \(x_k^* = 2 + k \frac{3}{n} = 2 + \frac{3k}{n}\text{.}\) The \(\frac{b-a}{n}\) terms becomes \(\frac{3}{n}\text{.}\) This term doesn’t involve \(k\text{,}\) so it can be factored out of the sum.
\begin{align*}
\amp = \lim_{n \rightarrow \infty} \frac{3}{n}
\sum_{k=1}^n \left( \left(2 + \frac{3k}{n} \right)^3 -
\left(2+ \frac{3k}{n} \right) \right)
\end{align*}
Then I’ll expand out the cubed binomial. After doing so, I’ll combine like terms (based on the power of \(n\) in the denominator).
\begin{align*}
\amp = \lim_{n \rightarrow \infty} \frac{3}{n}
\sum_{k=1}^n \left( 8 + \frac{36k}{n} + \frac{54k^2}{n^2}
+ \frac{27k^3}{n^3} - 2 - \frac{3k}{n} \right)\\
\amp = \lim_{n \rightarrow \infty} \frac{3}{n}
\sum_{k=1}^n \left( 6 + \frac{33k}{n} + \frac{54k^2}{n^2}
+ \frac{27k^3}{n^3} \right)\\
\amp = \lim_{n \rightarrow \infty} \frac{3}{n}
\left(
6 \sum_{k=1}^n 1 +
\frac{33}{n} \sum_{k=1}^n k +
\frac{54}{n^2} \sum_{k=1}^n k^2 +
\frac{27}{n^3} \sum_{k=1}^n k^3 \right)
\end{align*}
I evaluate all four of the sums, since each is one of the standard sums.
\begin{align*}
\amp = \lim_{n \rightarrow \infty} \frac{3}{n} \left( 6n +
\frac{33}{n} \frac{n(n+1)}{2} + \frac{54}{n^2}
\frac{n(n+1)(2n+1)}{6} + \frac{27}{n^3} \left(
\frac{n(n+1)}{2} \right)^2 \right)
\end{align*}
I simplify the expression by expanding the multiplication in each numerator. Then I can evaluate this as an asymptotic analysis limit. The asymptotic orders are the same in each term (and I can treat each term independently, since the limits splits up over addition), so the limit is the ratio of the leading terms in each piece of the sum. Adding the various pieces together completes the integral.
\begin{align*}
\amp = 18 + \frac{99}{2} + 54 + \frac{81}{4} = \frac{72+
199+ 216+ 81}{4} = \frac{567}{4}
\end{align*}
Subsection 8.2.2 Properties of the Definite Integral
The definite integral is linear.
\begin{align*}
\int_a^b (f(x) \pm g(x)) dx \amp = \int_a^b f(x) dx +
\int_a^b g(x) dx\\
\int_a^b c f(x) dx \amp = c \int_a^b f(x) dx
\end{align*}
To integrate a constant, I just calculate the area of a rectangle, since the area under a constant graph is a rectangle.
\begin{equation*}
\int_a^b c dx = c(b-a)
\end{equation*}
If, for some reason, I end up with reversed bounds, I can change them by introducing a negative sign.
\begin{equation*}
\int_a^b f(x) dx = - \int_b^a f(x) dx
\end{equation*}
If \(f\) is continuous on \([a,b]\) and \([b,c]\text{,}\) then the integral over the two pieces is equivalent to the integral over the whole interval \([a,c]\text{.}\)
\begin{equation*}
\int_a^bf(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx
\end{equation*}
