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Section 6.3 Calculating Integrals Using Accumulations

Overview.

We may be accustomed from past experience to expect that every mathematics problem is meant to have a nice simple formula. Definite integrals are an example of a mathematical calculation that is the opposite. If someone were to write down a randomly created definite integral, there would typically be no formula using elementary functions that calculate its value. However, there are some rules that we can use in particular situations.

This section focuses on some elementary formulas that are known for definite integrals. These formulas are called accumulation formulas. In light of the Fundamental Theorem of Calculus, the formulas also correspond to what is known as antiderivative. The most basic accumulation rule is for simple power functions.

Definite integrals satisfy a sum rule and constant multiple rule. These rules allow us to compute definite integrals whose rates of accumulation are formed from simpler rates through summation or through multiplication by a constant. Together with the elementary accumulation formulas that we learn, these rules allow us to compute a useful collection of definite integrals.

Subsection 6.3.1 Accumulation of Power Functions

The elementary power functions \(\mathrm{pow}_p(x) = x^p\) have a relatively simple associated accumulation function. One way to learn about the accumulation formula is to try the simplest examples and look to see if a pattern arises. This process is called forming a conjecture. Observed patterns are not always true patterns, so we will eventually need a more rigorous approach.

An accumulation function for a given integrable rate function \(f(x)\) can be constructed by calculating the definite integral from a particular constant starting point to the value of the independent variable. We write

\begin{equation*} A(x) = \int_a^x f(z) \, dz\text{,} \end{equation*}

where we use a variable of integration that is different from the \(x\) used as the independent variable.

The simplest rate function is a constant function, corresponding to a power \(p=0\text{,}\) \(f(x)=1\text{.}\) An elementary accumulation function would be the integral from \(0\) to \(x\text{,}\) corresponding to the signed area of a rectangle. Because the rectangle has a width height of \(f(x)=1\) and a width

\begin{equation*} A(x) = \int_0^x 1 \, dz = x\text{.} \end{equation*}

We know that

\begin{equation*} \int_a^b 1 \, dx = A(b) - A(a) = b-a\text{.} \end{equation*}

We can also easily find the definite integral of the identity function \(f(x)=x\text{,}\) which corresponds to a power \(p=1\text{.}\) Because the graph of \(y=f(z)=z\) is linear, the definite integral \(\displaystyle \int_0^x z \, dz\) corresponds to the signed area of a triangle. Since the triangle has a base \(x\) and a height \(f(x)=x\text{,}\) the definite integral that depends on end point \(x\) has a value

\begin{equation*} A(x) = \int_0^x z \, dz = \frac{1}{2}x^2\text{.} \end{equation*}

The definite integral from \(x=a\) to \(x=b\) can be computed as the change in accumulation,

\begin{equation*} \int_a^b z \, dz = A(b)- A(a) = \frac{1}{2}b^2 - \frac{1}{2}a^2\text{.} \end{equation*}

Higher powers become more complicated and required increasingly complicated results. For the quadratic power function \(f(x)=x^2\) with \(p=2\text{,}\) the graph takes the form of a parabola. The Greek philosopher and mathematician Archimedes used a method of exhaustion to determine the area enclosed by a parabola. In the context of an accumulation function, we have

\begin{equation*} A(x) = \int_0^x z^2 \, dz = \frac{1}{3}x^3 \text{.} \end{equation*}

For a cubic power function \(f(x) = x^3\) (\(p=3\)), an 11th century Arab mathematician Ibn al-Haytham during the Islamic Golden Age found a result that shows

\begin{equation*} A(x) = \int_0^x z^3 \, dz = \frac{1}{4}x^4 \text{.} \end{equation*}

We can recreate Archimedes's and Ibn al-Haytham's results using limits of Riemann sums.

Show that \(\displaystyle \int_0^x z^2 \, dz = \frac{1}{3}x^3\) and \(\displaystyle \int_0^x z^3 \, dz = \frac{1}{4}x^4\text{.}\)

Solution

For both integrals, the interval of integration goes from \(a=0\) to \(b=x\text{.}\) When we make a partition with \(n\) uniform subintervals, the width of each subinterval will be

\begin{equation*} \Delta x = \frac{x-0}{n} = \frac{x}{n}\text{.} \end{equation*}

In addition, the points in the partition will be defined by

\begin{equation*} x_k = \frac{kx}{n}, \quad k=0, 1, \ldots, n\text{.} \end{equation*}

The integral for \(f(z) = z^2\) uses values \(f(x_k) = \frac{k^2 x^2}{n^2}\) to give a Riemann sum

\begin{align*} \int_0^x z^2 \, dz \amp \approx \sum_{k=1}^{n} f(x_k) \Delta x \\ \amp = \sum_{k=1}^{n} \frac{k^2 x^2}{n^2} \cdot \frac{x}{n} \\ \amp = \sum_{k=1}^{n} \frac{k^2 x^3}{n^3} \\ \amp = \frac{x^3}{n^3} \cdot \sum_{k=1}^{n} k^2 \end{align*}

When we use the closed formula for the sum of squares and take a limit, we find

\begin{align*} \int_0^x z^2 \, dz \amp= \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x\\ \amp= \lim_{n \to \infty} \frac{x^3}{n^3} \sum_{k=1}^{n} k^2\\ \amp= \lim_{n \to \infty} \frac{x^3}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \\ \amp= \lim_{n \to \infty} \frac{x^3}{6} \cdot \frac{2n^3+3n^2+n}{n^3} \\ \amp= \lim_{n \to \infty} \frac{x^3}{6} \cdot (2 + \frac{3}{n} + \frac{1}{n^2}) \\ \amp= \frac{x^3}{6} \cdot (2 + 0 + 0) \\ \amp= \frac{x^3}{3} \end{align*}

The integral for \(f(z)=z^3\) is similar but with \(f(x_k) = \frac{k^3 x^3}{n^3}\) to give a Riemann sum

\begin{align*} \int_0^x z^3 \, dz \amp \approx \sum_{k=1}^{n} f(x_k) \Delta x \\ \amp = \sum_{k=1}^{n} \frac{k^3 x^3}{n^3} \cdot \frac{x}{n} \\ \amp = \sum_{k=1}^{n} \frac{k^3 x^4}{n^4} \\ \amp = \frac{x^4}{n^4} \cdot \sum_{k=1}^{n} k^3 \end{align*}

Applying the formula for the sum of cubes and a limit, we find

\begin{align*} \int_0^x z^3 \, dz \amp= \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x\\ \amp= \lim_{n \to \infty} \frac{x^4}{n^4} \sum_{k=1}^{n} k^3\\ \amp= \lim_{n \to \infty} \frac{x^4}{n^4} \cdot \frac{n^2(n+1)^2}{4} \\ \amp= \lim_{n \to \infty} \frac{x^4}{4} \cdot \frac{n^4+2n^3+n^2}{n^4} \\ \amp= \lim_{n \to \infty} \frac{x^4}{4} \cdot (1 + \frac{2}{n} + \frac{1}{n^2}) \\ \amp= \frac{x^4}{4} \cdot (1 + 0 + 0) \\ \amp= \frac{x^4}{4} \end{align*}

We start to look for possible patterns that might generalize these results. If we write the results side-by-side, we observe that each accumulation is also a power and with a pattern in the coefficient.

\begin{align*} \int_0^x z^0 \, dz \amp= x^1 \\ \int_0^x z^1 \, dz \amp= \frac{1}{2}x^2 \\ \int_0^x z^2 \, dz \amp= \frac{1}{3}x^3 \\ \int_0^x z^3 \, dz \amp= \frac{1}{4}x^4 \end{align*}

The power of the accumulation is always one value higher than the power of the rate function. The coefficient multiplying the accumulation is then the reciprocal of that higher power. We summarize our conjecture as a power rule for integration.

Our conjecture will ultimately be shown to be correct. However, we will need some additional tools before we know how to prove that it is true. Riemann sums will not help because we would need a summation formula for each possible case. We will need a more general concept of accumulation called an antiderivative. The power rule will, in fact, give us an antiderivative for power functions of any power \(p \ne -1\text{.}\)

Subsection 6.3.2 Accumulation Rules of Combination

Having found accumulation formulas for elementary power functions, we would like to find the definite integrals for more complicated functions. There are two basic rules of combination for definite integrals—the constant multiple and the sum rules.

Because \(f(x)\) is integrable, we know

\begin{equation*} \displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x = \int_a^b f(x) \, dx\text{.} \end{equation*}

If we then compute the Riemann sum for a rate \(k f(x)\text{,}\) each term is multiplied by \(k\text{.}\) Both the summation operator and the limit operator have a constant multiple rule, allowing the constant \(k\) to factor its way out.

\begin{align*} \int_a^b k \, f(x) \, dx \amp = \lim_{n \to \infty} \sum_{k=1}^{n} k f(x_k) \Delta x \\ \amp= \lim_{n \to \infty} k \sum_{k=1}^{n} f(x_k) \Delta x \\ \amp= k \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x \\ \amp= k \int_a^b f(x) \, dx \end{align*}

Geometrically, the constant multiple rule is about rescaling the graph of \(f(x)\) by a factor \(k\text{.}\) Because the width of the region is the same but every point has been stretched vertically by \(k\text{,}\) the signed area must be multiplied by \(k\text{.}\) The proof of the rule results because both limits and summation formulas have a constant multiple rule.

Because \(f(x)\) and \(g(x)\) are integrable, we know

\begin{equation*} \displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x = \int_a^b f(x) \, dx \end{equation*}

and

\begin{equation*} \displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} g(x_k) \Delta x = \int_a^b g(x) \, dx \end{equation*}

If we then compute the Riemann sum for rate formed by the sum \(f(x) + g(x)\text{,}\) then the sum rules for the summation operator and for the limit operator allow us to add the integrals.

\begin{align*} \int_a^b f(x) + g(x) \, dx \amp = \lim_{n \to \infty} \sum_{k=1}^{n} \left(f(x_k) + g(x_k)\right) \Delta x \\ \amp = \lim_{n \to \infty} \left(\sum_{k=1}^{n} f(x_k)\Delta x + \sum_{k=1}^{n} g(x_k) \Delta x \right) \\ \amp = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\Delta x + \lim_{n \to \infty} \sum_{k=1}^{n} g(x_k) \Delta x \\ \amp= \int_a^b f(x) \, dx + \int_a^b g(x) \, dx. \end{align*}

Geometrically, the sum rule is thinking of the Riemann sum rectangles with rate \(f(x) + g(x)\) and width \(\Delta x\) as being formed by two rectangles of width \(\Delta x\text{.}\) One rectangle has height \(f(x)\) and the other has height \(g(x)\text{.}\) The total area is the sum \(f(x) \Delta x + g(x) \Delta x\text{.}\) When we add all of the increments, we can group all of the parts from \(f(x) \Delta x\) to give a sum matching the integral \(\displaystyle \int_a^b f(x) \,dx\text{.}\) The sum of the other terms give \(\displaystyle \int_a^b g(x) \,dx\text{.}\)

With our two rules of arithmetic and the power rule for accumulations, we can compute the definite integrals of polynomials.

Example 6.3.5.

Find \(\displaystyle \int_2^4 3x^2 - 4 \, dx\text{.}\)

Solution

We start by looking at the integrand. The rate function \(f(x) = 3x^2-4\) is a linear combination of the elementary powers \(x^2\) and \(x^0 = 1\text{,}\) \(f(x) = 3 \cdot x^2 + -4 \cdot 1\text{.}\) The power rule gives us accumulation formulas for the powers, so we know

\begin{align*} \int_2^4 x^2 \, dx \amp= \frac{1}{3}(4^3) - \frac{1}{3}(2^3)\\ \amp= \frac{64}{3}-\frac{8}{3} = \frac{56}{3} \\ \int_2^4 1 \, dx \amp= 4-2 = 2 \end{align*}

Using the sum and constant multiple rules, we find

\begin{align*} \int_2^4 3x^2 - 4 \, dx \amp= 3 \int_2^4 x^2 \, dx + -4 \int_2^4 1 \, dx\\ \amp= 3 \cdot \frac{56}{3} + -4 \cdot 2 \\ \amp= 56-8 = 48 \end{align*}

In the previous example, we used the constant multiple and sum rule to break a definite integral of a linear combination into a corresponding linear combination of the values of the component definite integrals. However, it is usually more convenient to create an accumulation function that is a linear combination of the corresponding accumulations of the component rates. That is, a rate function \(f(x) = 3x^2 - 4\) must have a corresponding accumulation function

\begin{equation*} A(x) = 3 \cdot \frac{1}{3} x^3 + -4 \cdot x = x^3 - 4x, \end{equation*}

based on the accumulation functions for the rates \(x^2\) and \(1\text{.}\) The definite integral is then just the change in this new accumulation function.

Example 6.3.6.

Find \(\displaystyle \int_{-2}^{2} 2x^2 - 3x + 4 \, dx\text{.}\)

Solution

The rate function in then integrand is \(f(x) = 2x^2 - 3x + 4\text{.}\) The accumulations for elementary powers are \(\frac{1}{3}x^3\) for \(x^2\text{,}\) \(\frac{1}{2}x^2\) for \(x\text{,}\) and \(x\) for \(1\text{.}\) Consequently, the accumulation for \(f(x)\) is

\begin{equation*} A(x) = 2 \cdot \frac{1}{3}x^3 + -3 \cdot \frac{1}{2} x^2 + 4 \cdot x\text{.} \end{equation*}

The definite integral of the rate will equal the change in the accumulation,

\begin{align*} \int_{-2}^{2} 2x^2 - 3x + 4 \, dx \amp= A(2) - A(-2) \\ \amp= \left(\frac{2}{3}(2)^3 - \frac{3}{2} (2)^2 + 4(2) \right) - \left(\frac{2}{3}(-2)^3 - \frac{3}{2} (-2)^2 + 4(-2) \right)\\ \amp= \left(\frac{16}{3} - 6 + 8 \right) - \left(\frac{-16}{3} - 6 - 8 \right)\\ \amp= \frac{32}{3} + 16 = \frac{32+48}{3} = \frac{80}{3} \end{align*}

Subsection 6.3.3 Summary

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