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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 4.2 The Definite Integral

Subsection 4.2.1 Before Class

https://mymedia.ou.edu/media/4.2-1/1_t1bcahd5
Figure 38. Pre-Class Video 1

Subsubsection 4.2.1.1 Definition of the Definite Integral

Definition 4.2.1.
Let \(f\) be a function defined for \(a\leq x\leq b\text{,}\) and divide \([a,b]\) into \(n\) subintervals of width \(\Delta x = \dfrac{b-a}{n}\text{.}\) Let
\begin{equation*} a=x_0,x_1,x_2,\cdots,x_{n-1},x_n=b \end{equation*}
be the endpoints of these subintervals, and let \(x_i^*\) be any sample points in these subintervals, so that \(x_i^*\) lies in the \(i\)th subinterval \([x_{i-1},x_i]\text{.}\) Then, the definite integral of \(f\) from \(a\) to \(b\) is given by
\begin{equation*} \ds \int_a^b f(x)\, dx = \sum_{n=1}^\infty f(x_i^*)\Delta x \end{equation*}
provided that this limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that \(f\) is integrable on \([a,b]\text{.}\) The sum \(\ds \sum_{n=1}^\infty f(x_i^*)\Delta x\) is called a Riemann sum.
Terminology.
Given the definite integral \(\ds \int_a^b f(x)\, dx\text{,}\) we have some terminology:
  • \(a\) is called the lower bound/lower limit of integration
  • \(b\) is called the upper bound/upper limit of integration
  • \(\ds \int\) is called an integral sign
  • We use \(dx\) to denote an infinitesimal change in x
Interpretation of the Integral.
The definite integral \(\ds \int_a^b f(x)\, dx\) can be interpreted as
  • The total area between \(f(x)\) and the \(x-\)axis if \(f(x)\geq 0\) on \([a,b]\)
  • The net area between \(f(x)\) and the \(x-\)axis if \(f(x)\geq 0\) and \(f(x) < 0\) on \([a,b]\)
Here are some pictures to help with the interpretations:
This is the graph of \(f(x) = 1\) on the interval \([0,1]\)
Figure 39. Here, the net area and total area are both 1, because the function is strictly positive on the interval.
This is the graph of \(f(x) = 1\) on the interval \([0,1]\)
Figure 40. Here, the net area is 0 but the total area is 2 because the function takes on positive and negative values in the interval.
Example 4.2.4.
Express \(\ds \lim_{n\to\infty} \sum_{i=1}^\infty (x_i^3-2x_i^2+1-x_i\cos(x_i))\Delta x\) as an integral. Use proper notation.
Solution.
\(\ds \int_a^b x^3-2x^2 + 1 - x\cos x\, dx\)
Example 4.2.5.
  1. The greatest integer function is integrable on any finite-length interval, but not necessarily differentiable on the same interval. Why is this consistent with the above information?
  2. Compute \(\ds \int_0^5 f(x)\, dx\text{,}\) where \(f(x)\) is the greatest integer function, using geometry.
Solution.
  1. This is because the number of discontinuities on a finite-length interval is finite, so from the above theorem, the function is integrable.
  2. \(\displaystyle \ds \int_0^5 f(x)\, dx = 10\)

Subsection 4.2.2 Pre-Class Activities

Example 4.2.6.

The graph of \(\sin x\) on \([-3\pi,3\pi]\text{,}\) with shading from \(x=0\) to \(x=\pi/2\)
The graph of \(\sin x\) on \([-3\pi,3\pi]\text{,}\) with shading from \(x=0\) to \(x=\pi/2\)
The shaded area on the graphs of \(\sin x\) and \(\cos x\) are both exactly 1.
  1. Compute the net area of \(\sin x\) on the interval \([0,2\pi]\)
  2. Compute the total area of \(\sin x\) on the interval \([0,2\pi]\)
  3. Compute the net area of \(\cos x\) on the interval \(\left[-\dfrac{3\pi}{2},\pi\right]\)
  4. Compute the total area of \(\cos x\) on the interval \(\left[-\dfrac{3\pi}{2},\pi\right]\)
  5. Compute the net area of \(\sin x\) on the intervals \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) and \([-\pi,\pi]\text{.}\) What conjecture can you make about the net area of odd functions over symmetric intervals?
  6. Compute the net area of \(\cos x\) on the intervals \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) and \([-\pi,\pi]\text{.}\) What conjecture can you make about the net area of odd functions over symmetric intervals?
Solution.
  1. 0
  2. 2
  3. \(\displaystyle -1\)
  4. 5
  5. Both are 0. Conjecture is left to student.
  6. First interval is 2, second interval is 0. Conjecture left to student.

Subsection 4.2.3 In Class

Subsubsection 4.2.3.1 Evaluating Integrals

Example 4.2.7.
Use the definition of the definite integral to show that \(\ds \int_2^4 x\, dx = 6\)
Solution.
First note that
\begin{equation*} \Delta x = \dfrac{b-a}{n} = \dfrac{2}{n} \end{equation*}
For \(x_i\text{,}\) we have
\begin{equation*} x_i = a + i\Delta x = 2 + \dfrac{2i}{n} \end{equation*}
This means that
\begin{equation*} f(x_i) = 2 + \dfrac{2i}{n} \end{equation*}
So that the area of a single rectangle is given by
\begin{equation*} f(x_i)\Delta x = \lrpar{2+\dfrac{2i}{n}}\cdot \dfrac{2}{n} = \dfrac{4}{n} + \dfrac{4i}{n^2} \end{equation*}
And the sum of a finite number of rectangles is
\begin{equation*} \ds \sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \dfrac{4}{n} + \dfrac{4i}{n^2} = 4 + \dfrac{2n(n+1)}{n^2} \end{equation*}
after using the helpful formulas given below. Now, the integral can be computed by taking the limit:
\begin{equation*} \ds \int_2^4 x\, dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x = 4 + 2 = 6 \end{equation*}
So the definite integral is 6.
Helpful Formulas Rules for Sigma Notation
\(\ds \sum_{i=1}^n i\) \(\dfrac{n(n+1)}{2}\) \(\ds \sum_{i=1}^n c\) \(cn\)
\(\ds \sum_{i=1}^n i^2\) \(\dfrac{n(n+1)(2n+1)}{6}\) \(\ds \sum_{i=1}^n ca_i\) \(\ds c\sum_{i=1}^n a_i\)
\(\ds \sum_{i=1}^n i^3\) \(\lrpar{\dfrac{n(n+1)}{2}}^2\) \(\ds \sum_{i=1}^n (a_i\pm b_i)\) \(\ds \sum_{i=1}^n a_i \pm \sum_{i=1}^n b_i\)
Example 4.2.8.
Evaluate \(\ds \int_0^3 (x^3-2x)\, dx\) using the definition of the definite integral.
Solution.
\(\dfrac{45}{4}\)
Example 4.2.9.
Set up an expression for \(\ds \int_2^5 x^4\, dx\) as a limit of sums.
Solution.
\(\ds \lim_{n\to\infty} \sum_{i=1}^n \lrpar{2 + \dfrac{3i}{n}}^4\cdot \dfrac{3}{n}\)
Example 4.2.10.
The graph of \(f\) is given below.
Evaluate each integral by interpreting it in terms of areas.
  1. \(\displaystyle \ds \int_0^2 f(x)\, dx\)
  2. \(\displaystyle \ds \int_0^5 f(x)\, dx\)
  3. \(\displaystyle \ds \int_5^7 f(x)\, dx\)
  4. \(\displaystyle \ds \int_0^9 f(x)\, dx\)
Solution.
  1. 4
  2. 10
  3. \(\displaystyle -3\)
  4. 2

Subsubsection 4.2.3.2 The Midpoint Rule

Instead of choosing our sample points to be left or right endpoints, we can choose any point inside the subinterval; a standout is the midpoint of the interval.
Definition 4.2.11. Midpoint Rule.
\begin{equation*} \ds \int_a^b f(x)\, dx \approx \sum_{i=1}^n f(\overline{x_i})\Delta x \end{equation*}
where \(\Delta x = \dfrac{b-a}{n}\) and \(\overline{x_i} = \dfrac{x_{i-1}+x_i}{2}\) is the midpoint of the interval \([x_{i-1},x_i]\text{.}\)
Example 4.2.12.
Use the Midpoint Rule with \(n=5\) to approximate \(\ds \int_1^2 \dfrac{1}{x}\, dx\)
Solution.
\(M_5 = 0.69191\)

Subsubsection 4.2.3.3 Properties of the Definite Integral

The following are useful properties when working with the definite integral:
Example 4.2.14.
If \(\ds \int_0^6 f(x)\, dx = 10\) and \(\ds \int_0^8 f(x)\, dx = 4\text{,}\) what is \(\ds \int_6^8 f(x)\, dx\text{?}\)
Solution.
\(\ds \int_6^8 f(x)\, dx = -6\)
Example 4.2.15.
Compute \(\ds \int_1^2 (4+2x^2)\, dx\)
Solution.
\(\dfrac{26}{3}\)
Example 4.2.16.
For any \(a,b\text{,}\) find the value of \(\ds \int_a^b x\, dx\)
Solution.
\(\ds \int_a^b x\, dx = \dfrac{b^2-a^2}{2}\)
Example 4.2.17.
Evaluate the integrals by interpreting them in terms of area:
  1. \(\displaystyle \ds \int_0^9 \lrpar{\dfrac{2}{3}x-2}\, dx\)
  2. \(\displaystyle \ds \int_{-6}^6 (x-\sqrt{36-x^2})\, dx\)
  3. \(\displaystyle \ds \int_0^1 |2x-1|\, dx\)
Solution.
  1. \(\displaystyle \ds \int_0^9 \lrpar{\dfrac{2}{3}x-2}\, dx = 9\)
  2. \(\displaystyle \ds \int_{-6}^6 (x-\sqrt{36-x^2})\, dx = -18\pi\)
  3. \(\displaystyle \ds \int_0^1 |2x-1|\, dx = \dfrac{1}{2}\)

Subsection 4.2.4 After Class Activities

Example 4.2.18.

Use the definition of the integral to find \(\ds \int_1^4 (x^2-4x+2)\, dx\)
Solution.
\(\ds \int_1^4 (x^2-4x+2)\, dx = -3\)

Example 4.2.19.

Use the Midpoint Rule to estimate \(\ds \int_0^1 \sqrt{x^3+1}\, dx\) to four decimal places, using \(n=5\) rectangles.
Solution.
\(M_5 = 1.1097\)

Example 4.2.20.

If \(\ds \int_0^\pi \sin^4x\, dx = \dfrac{3\pi}{8}\text{,}\) what is \(\ds \int_\pi^0 \sin^4\, dx\text{?}\) Why?
Solution.
\(\ds \int_\pi^0 \sin^4\, dx = -\dfrac{3\pi}{8}\) from the properties of integrals.
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