Warming Up

Section 1.1 Warming Up

This second calculus course starts where your previous calculus course finished.

The expectation is that you know the derivatives of basic functions (power functions, trigonometric functions, and exponential and logarithmic functions), and that you have a good working knowledge of the fundamental rules of differentiation (product rule, quotient rule, and chain rule).

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In this section, we briefly review the basic concepts of antiderivatives and indefinite integrals, definite integrals, and the method of substitution. It is essential that you understand these concepts, as they form the foundation for this course.

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When you work through this review, you may discover that you are a little bit rusty. That is normal, especially if it has been a few years since you completed the previous calculus course. In that case, be sure to take the time to review the topics carefully by working through exercises on these topics (these topics are generic for any calculus sequence; any calculus textbook will suffice for this purpose).

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Subsection 1.1.1 Antiderivatives and Indefinite Integrals

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Recall the following definition:

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Definition 1.1.1.

A function

$F$ is called an antiderivative of

$f$ on an interval

$I$ if

$F'(x) = f(x)$ for all

$x$ in

$I\text{.}$

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Also recall that if

$F$ is an antiderivative of

$f$ on an interval

$I\text{,}$ then so is

$F(x)+C$ where

$C$ is an arbitrary constant. We refer to

$C$ as the constant of integration.

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Checkpoint 1.1.2. Some Basic Antiderivatives.

Determine the most general antiderivatives of $\displaystyle f(x) = x^4\text{,}$ $g(x) = \sin(x)\text{,}$ and $\displaystyle h(x) = \frac{1}{x}\text{.}$

Solution

$\displaystyle F(x) = \frac{1}{5} x^5 + C\text{,}$ since $\displaystyle F'(x) = \frac{1}{5} \cdot 5 x^4 = x^4 = f(x)\text{.}$

$\displaystyle G(x) = - \cos(x) + C\text{,}$ since $\displaystyle G'(x) = - (-\sin(x)) = \sin(x) = g(x)\text{.}$

$\displaystyle H(x) = \ln|x| + C\text{,}$ since $\displaystyle H'(x) = \frac{1}{x} = h(x)\text{.}$

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The above notation with capital letters is rather cumbersome. What if the given function is

$F(x)$ and you are asked to find its antiderivative, for example? Recall that we have generic notation for antiderivatives, namely the indefinite integral:

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$\displaystyle \int f(x) \ dx = F(x) \ \ \ {\rm means} \ \ \ F'(x) = f(x).$

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Example 1.1.3. Some Basic Indefinite Integrals.

Thus, we can write the results of the previous checkpoint as follows:

$\displaystyle \displaystyle \int x^4 \ dx = \frac{1}{5} x^5 + C$
$\displaystyle \displaystyle \int \sin(x) \ dx = - \cos(x) + C$
$\displaystyle \displaystyle \int \frac{dx}{x} = \ln|x| + C$

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Activity 1.1.1.

Take some time and create as complete a list of basic indefinite integrals that you can. Review the basic derivatives from your previous calculus course, and write the corresponding indefinite integrals.

For example, a basic rule of differentiation is the power rule, namely $\displaystyle \frac{d}{dx} x^n = n x^{n-1}\text{,}$ which we can rewrite as $\displaystyle \frac{d}{dx} \frac{1}{n} x^n = x^{n-1}$ provided $n \ne 0\text{,}$ or $\int x^n \ dx = \frac{x^{n+1}}{n+1} + C \ \ \ (n \ne -1).$ You should be able to write at least 10 basic indefinite integrals. We will complete the list in class.

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Remark 1.1.4.

We will see later in the course that the constant of integration is really important, so it is a good idea to get into the habit of always writing the constant of integration, except when you know for sure that it is irrelevant.

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Subsection 1.1.2 Definite Integrals

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In general (if

$f$ is continuous on

$[a,b]$ or if

$f$ only has a finite number of jump discontinuities), the definite integral

$\int_a^b f(x) \ dx$ evaluates to a number.

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Question 1.1.5. Interpretation of the Definite Integral.

Do you remember how to interpret a definite integral in terms of area under the curve $f(x)$ between $x=a$ and $x=b\text{?}$ What if $f(x)$ lies entirely above the $x$-axis? Or entirely below the $x$-axis? Or if it lies partly above and partly below the $x$-axis?

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Subsection 1.1.3 The Evaluation Theorem for Definite Integrals

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The Evaluation Theorem provides an efficient method to determine the value of a definite integral

$\int_a^b f(x) \ dx$ when it is possible to determine an antiderivative of

$f\text{.}$

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Theorem 1.1.6. The Evaluation Theorem.

$f$ is continuous on the interval

$[a, b]\text{,}$ then

$\int_a^b f(x) \ dx = \left. F(x) \right|_a^b = F(b) - F(a),$ where

$F$ is any antiderivative of

$f\text{,}$ that is,

$F' = f\text{.}$

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Checkpoint 1.1.7. An Application of the Evaluation Theorem.

Evaluate $\displaystyle \int_{0}^{\pi/4} \sec^2(x) \ dx.$

Hint

Recall that $\displaystyle \frac{d}{dx} \tan(x) = \sec^2(x).$

Answer

$\displaystyle \int_{0}^{\pi/4} \sec^2(x) \ dx = 1\text{.}$

Solution

\begin{alignat*}{1} \displaystyle \int_{0}^{\pi/4} \sec^2(x) \ dx &= \left. \tan(x) \right|_0^{\pi/4} \\ &= \tan(\pi/4) - \tan(0) \\ &= 1 - 0 \\ &= 1 \end{alignat*}

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Remark 1.1.8.

Note that it is not necessary to include the constant of integration in the evaluation of a definite integral. Why not? Think about it!

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If you wish a refresher on the use of the Evaluation Theorem to evaluate indefinite and definite integrals, you may find it helpful to review the following video (from a first course on calculus such as MATH 114 or 134).

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Figure 1.1.9. A video with various examples using the Evaluation Theorem.

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Subsection 1.1.4 Substitution

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While we can readily write down

$\int \sqrt{x} \ dx = \int x^{1/2} \ dx = \frac{1}{3/2} x^{3/2} + C = \frac{2}{3} x \sqrt{x} + C,$ what about

$\displaystyle \int \sqrt{ 3x+4 } \ dx\text{?}$

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You likely expect the answer to be of the form

$A (3x+4)^{3/2} + C$ and, working backwards by taking the derivative of this result using the chain rule, you then can determine the value of

$A\text{.}$

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Checkpoint 1.1.10. Working Backwards.

For $\displaystyle \int \sqrt{ 3x+4 } \ dx = A (3x+4)^{3/2} + C\text{,}$ determine the value of $A\text{.}$

Hint

Don't forget to use the chain rule!

Answer

$\displaystyle A = \frac{2}{9}.$

Solution

Using the chain rule, we have $$ \frac{d}{dx} \left[ A (3x+4)^{3/2} + C \right] = A \cdot \frac{3}{2} ( 3x+4 )^{1/2} (3) = \frac{9}{2} A \sqrt{ 3x+4 }. $$ Since the original integrand was simply $\sqrt{ 3x+4 }\text{,}$ it must be true that $\displaystyle \frac{9}{2} A = 1\text{,}$ that is, $\displaystyle A = \frac{2}{9}\text{.}$

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But this is rather cumbersome and ad hoc, and unlikely to work when the integrand is more complicated. Substitution is a method to 'undo' the chain rule, and is one of the most important methods to evaluate integrals when inspection fails.

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Recall that the chain rule states

$\frac{d}{dx} f( g(x) ) = f'( g(x) ) g'(x).$ If we rewrite the chain rule in integral form, we obtain

$\int f'( g(x) ) g'(x) \ dx = f( g(x) ) + C.$

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If we let

$u=g(x)\text{,}$ then

$\displaystyle \frac{du}{dx} = g'(x)\text{,}$ or

$du = g'(x) dx$ in differential form, then substituting

$u$ and

$du$ into the chain rule in integral form above gives

$\int f'( g(x) ) g'(x) \ dx = \int f'(u) \ du = f(u) + C = f( g(x) ) + C.$ While we may not be able to readily go from the left-hand side to the right-hand side, the idea of substitution is to simplify the integrand in the intermediate step, thereby allowing us to make progress.

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Checkpoint 1.1.11. Substitution: Starter Example.

Use substitution to evaluate $\displaystyle \int \sqrt{ 3x+4 } \ dx\text{.}$

Hint

Use the substitution $u=3x+4; du = 3 dx\text{.}$

Solution

Let $u=3x+4\text{.}$ Then $du = 3 dx$ or $\displaystyle dx = \frac{1}{3} du\text{,}$ and we obtain

\begin{alignat*}{1} \int \sqrt{ 3x+4 } \ dx &= \int \sqrt{u} \frac{1}{3} \ du \\ &= \frac{1}{3} \int \sqrt{u} \ du \\ &= \frac{1}{3} \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{9} (3x+4)^{3/2} + C \end{alignat*}

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Checkpoint 1.1.12. Substitution: Medium Difficulty.

Evaluate $\displaystyle \int \frac{ \ln(x) }{x} \ dx\text{.}$

Hint

Use the substitution $\displaystyle u=\ln(x); du = \frac{1}{x} dx\text{.}$

Answer

$\displaystyle \int \frac{ \ln(x) }{x} \ dx = \frac{1}{2} \ln^2(x) + C\text{.}$

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Checkpoint 1.1.13. Substitution: More Challenging.

Evaluate $\displaystyle \int x \sqrt{ x+2 } \ dx\text{.}$

Hint

Use the substitution $u=x+2\text{,}$ and note that you can rewrite this as $x = u-2\text{.}$

Answer

$\displaystyle \int x \sqrt{ x+2 } \ dx = \frac{2}{5} (x+2)^{5/2} - \frac{4}{3} (x+2)^{3/2} + C\text{.}$

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If you wish a refresher on the method of substitution to evaluate indefinite and definite integrals, you may find it helpful to review the following videos.

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Figure 1.1.14. A review video on the substitution rule.

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Figure 1.1.15. A review video on the substitution rule for definite integrals.

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Subsection 1.1.5 Summary

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Evaluation Theorem.

If $f$ is continuous on the interval $[a, b]\text{,}$ then $\int_a^b f(x) \ dx = \left. F(x) \right|_a^b = F(b) - F(a),$ where $F$ is any antiderivative of $f\text{,}$ that is, $F' = f\text{.}$

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Method of Substitution.

Substitution is a method to reverse the chain rule in the evaluation of integrals.
- Substitution for indefinite integrals:
  
  If $u = g(x)\text{,}$ then $\int f( g(x) ) g'(x) \ dx = \int f(u) \ du.$
- Substitution for definite integrals:
  
  If $g'$ is continuous on $[a, b]$ and $f$ is continuous on the range of $u = g(x)\text{,}$ then $\int_a^b f( g(x) ) g'(x) \ dx = \int_{g(a)}^{g(b)} f(u) \ du.$

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Subsection 1.1.6 Further Study

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The notes presented above only serve as a brief review of some key material from the first calculus course. Further study of this material may be required. This includes reviewing the lecture notes, and diligently working through the homework problems.

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As you study, you should reflect on the following learning outcomes, and critically assess where you are on the path to achieving these learning outcomes:

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Learning Outcomes

Apply the substitution rule to evaluate indefinite and definite integrals.

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The following references provide a good start for review and further study:

Learning Outcome	Video	Textbook Section
1	10.E1 from MATH 134 12.E4 from MATH 134	5.4