Section 6.3 Calculating Integrals Using Accumulations
ΒΆOverview.
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 powp(x)=xp 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 writeCheckpoint 6.3.1.
Show that \(\displaystyle \int_0^x z^2 \, dz = \frac{1}{3}x^3\) and \(\displaystyle \int_0^x z^3 \, dz = \frac{1}{4}x^4\text{.}\)
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
In addition, the points in the partition will be defined by
The integral for \(f(z) = z^2\) uses values \(f(x_k) = \frac{k^2 x^2}{n^2}\) to give a Riemann sum
When we use the closed formula for the sum of squares and take a limit, we find
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
Applying the formula for the sum of cubes and a limit, we find
Conjecture 6.3.2.
For any power p>0, we will have
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 integralsβthe constant multiple and the sum rules.Theorem 6.3.3.
If f(x) is integrable on an interval containing a,b and k is a constant, then
Proof.
Because \(f(x)\) is integrable, we know
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.
Theorem 6.3.4.
If f(x) and g(x) are each integrable on an interval containing a,b, then
Proof.
Because \(f(x)\) and \(g(x)\) are integrable, we know
and
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.
Example 6.3.5.
Find β«423x2β4dx.
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
Using the sum and constant multiple rules, we find
Example 6.3.6.
Find β«2β22x2β3x+4dx.
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
The definite integral of the rate will equal the change in the accumulation,