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Math 104: Calculus I – Notes
Enrolling in AP Calculus comes with the understanding that you will take the AP exam in May. The 2019 test will be given May 5, 2020If you do not plan on taking the AP Exam, we must have a conversation about it first. Mathematics 104—Calculus, Part I (4h, 1 CU) Course Description: Brief review of High School Calculus, methods and applications of integration, infinite series, Taylor's theorem, first order ordinary differential equations. Use of symbolic manipulation and graphics software in Calculus. Note: This course uses Maple®. This chapter reviews the basic ideas you need to start calculus.The topics include the real number system, Cartesian coordinates in the plane, straight lines, parabolas, circles, functions, and trigonometry. We also discuss the use of graphing.
Section 004 - Spring 2014
- Syllabus
| Skeleton Notes | Complete Notes | Title | More |
| Remainder 10.6, 10.9 | Remainder 10.6/10.9 | Series Estimation & Remainder | |
| Sections 10.8-10.10 | Sections 10.8-10.10 | Taylor (and Maclaurin) Series | |
| Section 10.7 | Section 10.7 | Power Series Introduction | |
| Section 10.6 | Section 10.6 | Alt. Series Test and Abs. Conv. | Conv. Tests |
| Section 10.5 | Section 10.5 | The Ratio and Root Tests | |
| Section 10.4 | Section 10.4 | The Comparison Tests | |
| Section 10.3 | Section 10.3 | The Integral Test | |
| Section 10.2 | Section 10.2 | Introduction to Series | |
| Section 10.1 | Section 10.1 | Sequences | |
| Section 9.2 | Section 9.2 | Linear Differential Equations | |
| Section 7.2 Pt 1Pt 2 | Section 7.2 | Separable Differential Equations | |
| Section 8.8 | Section 8.8 | Probability and Calculus | Odd Ans. |
| Section 8.7 Pt. 1Pt. 2 | Section 8.7 | Improper Integrals | L'Hopital |
| Section 8.4 Pt. 1Pt. 2 | Section 8.4 | Partial Fraction Decomposition | |
| Section 8.3 Pt. 1Pt. 2 | Section 8.3 | Trig. Substitution | |
| Section 8.2 Pt. 1Pt. 2 | Section 8.2 | Integrating Trig. Powers | |
| Section 8.1 Pt. 1Pt. 2 | Section 8.1 | Integration By Parts | |
| Section 6.6 | Section 6.6 | Center of Mass | |
| Section 6.4 | Section 6.4 | Surface Area of Revolution | |
| Section 6.3 | Section 6.3 | Arc Length | |
| Section 6.2 | Section 6.2 | Volumes Using Cylindrical Shells | |
| Section 6.1 | Section 6.1 | Volumes Using Cross-Sections | disk/washer |
| Review | |||
| Calc I Review | Calc I Review | Limit, Derivative, and Integral | |
| Area b/w Curves | Area b/w Curves Video Example | ||
| U-substitution | Graphs you should know |
Print out the skeleton notes before class and bring
them to class so that you don't have to write down
everything said in class. If you miss anything, the
complete notes will be posted after class.
5 Steps to a 5 AP Calculus AB & BC, 2012-2013 Edition (2011)
STEP 4. Review the Knowledge You Need to Score High
Chapter 10. Integration
IN THIS CHAPTER
Summary: On the AP Calculus exams, you will be asked to evaluate integrals of various functions. In this chapter, you will learn several methods of evaluating integrals including U-Substitution, Integration by Parts, and Integration by Partial Fractions. Also, you will be given a list of common integration and differentiation formulas, and a comprehensive set of practice problems. It is important that you work out these problems and check your solutions with the given explanations.
Key Ideas
Evaluating Integrals of Algebraic Functions
Integration Formulas
U-Substitution Method Involving Algebraic Functions
U-Substitution Method Involving Trigonometric Functions
U-Substitution Method Involving Inverse Trigonometric Functions
U-Substitution Method Involving Logarithmic and Exponential Functions
Integration by Parts
Integration by Partial Fractions
10.1 Evaluating Basic Integrals
Main Concepts: Antiderivatives and Integration Formulas, Evaluating Integrals
• Answer all parts of a question from Section II even if you think your answer to an earlier part of the question might not be correct. Also, if you do not know the answer to part one of a question, and you need it to answer part two, just make it up and continue.
Antiderivatives and Integration Formulas
Definition: A function F is an antiderivative of another function f if F′(x) = f(x) for all x in some open interval. Any two antiderivatives of f differ by an additive constant C. We denote the set of antiderivatives of f by ∫ f(x)dx, called the indefinite integral of f.
Integration Rules:
More Integration Formulas:
16. ∫ tan x dx = ln |sec x| + C or − ln |cos x | + C
17. ∫ cot x dx = ln |sin x| + C or − ln |csc x | + C
18. ∫ sec x dx = ln |sec x + tan x| + C
19. ∫ csc x dx = ln |csc x − cot x| + C
20. ∫ ln x dx = x ln |x| − x + C
21.
22.
23.
24.
Note: After evaluating an integral, always check the result by taking the derivative of the answer (i.e., taking the derivative of the antiderivative).
• Remember that the volume of a right-circular cone is where r is the radius of the base and h is the height of the cone.
Evaluating Integrals
Example 1
Example 2
Example 3
If , and the point (0, −1) lies on the graph of y, find y.
Since , then y is an antiderivative of . Thus, y = ∫ (3x2 + 2) dx = x3 + 2x + C. The point (0, −1) is on the graph of y. Thus, y = x3 + 2x + C becomes −1 = 03 + 2(0) + C or C = −1. Therefore, y = x3 + 2x − 1.
Example 4
Example 5
Example 6
Example 7
Example 8
Evaluate ∫ (4 cos x − cot x) dx.
∫ (4 cos x − cot x) dx = 4 sin x − ln |sin x| + C.
Example 9
Example 10
Example 11
Example 12
Example 13
Reminder: You can always check the result by taking the derivative of the answer.
• Be familiar with the instructions for the different parts of the exam before the day of exam. Review the instructions in the practice tests provided at the end of this book.
10.2 Integration by U-Substitution
Main Concepts:
The U-Substitution Method, U-Substitution and Algebraic Functions, U-Substitution and Trigonometric Functions, U-Substitution and Inverse Trigonometric Functions, U-Substitution and Logarithmic and Exponential Functions
The U-Substitution Method
The Chain Rule for Differentiation
, where F′ = f
The Integral of a Composite Function
If f(g(x)) and f′ are continuous and F′ = f, then
∫ f(g(x))g′(x)dx = F(g(x)) + C.
Making a U-Substitution
Let u = g(x), then du = g′(x)dx
∫ f(g(x))g″(x)dx = ∫ f(u)du = F(u) + C = F(g(x)) + C.
Procedure for Making a U-Substitution
Steps:
1. Given f(g(x)); let u = g(x).
2. Differentiate: du = g′(x)dx.
3. Rewrite the integral in terms of u.
4. Evaluate the integral.
5. Replace u by g(x).
6. Check your result by taking the derivative of the answer.
U-Substitution and Algebraic Functions
Another Form of the Integral of a Composite Function
If f is a differentiable function, then
Making a U-Substitution
Let u = f(x); then du = f′(x)dx.
Example 1
Evaluate .
Step 1. Let u = x + 1; then x = u − 1.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check:
Example 2
Evaluate .
Step 1. Let u = x − 2; then x = u + 2.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace: .
Step 6. Differentiate and Check:
Example 3
Evaluate .
Step 1. Let u = 2x − 5.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 4
Evaluate .
Step 1. Let u = x3 − 8.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
U-Substitution and Trigonometric Functions
Example 1
Evaluate .
Step 1. Let u = 4x.
Step 2. Differentiate: du = 4 dx or .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 2
Evaluate .
Step 1. Let u = tan x.
Step 2. Differentiate: du = sec2x dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: 2(tan x)3/2 + C or 2 tan3/2x + C.
Step 6. Differentiate and Check: .
Example 3
Evaluate .
Step 1. Let u = x3.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
• Remember that the area of a semi-circle is . Do not forget the . If the cross sections of a solid are semi-circles, the integral for the volume of the solid will involve which is .
U-Substitution and Inverse Trigonometric Functions
Example 1
Evaluate .
Step 1. Let u = 2x.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check:
Example 2
Evaluate .
Step 1. Rewrite:
.
Let u = x + 1.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check:
• If the problem gives you that the diameter of a sphere is 6 and you are using formulas such as or s = 4π r2, do not forget that r = 3.
U-Substitution and Logarithmic and Exponential Functions
Example 1
Evaluate .
Step 1. Let u = x4 − 1.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 2
Evaluate .
Step 1. Let u = cos x + 1.
Step 2. Differentiate: du = −sin xdx ⇒ − du = sin xdx.
Step 3. Rewrite: .
Step 4. Integrate: − ln |u| + C.
Step 5. Replace u: − ln |cos x + 1| + C.
Step 6. Differentiate and Check: .
Example 3
Evaluate .
Step 1. Rewrite ; by dividing (x2 + 3) by (x − 1).
Let u = x − 1.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: 4 ln |u| + C.
Step 5. Replace u: 4 ln |x − 1| + C.
.
Step 6. Differentiate and Check:
.
Example 4
Evaluate .
Step 1. Let u = ln x.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 5
Evaluate .
Step 1. Let u = 2x − 5.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 6
Evaluate .
Step 1. Let u = ex + 1.
Step 2. Differentiate: du = exdx.
Step 3. Rewrite: .
Step 4. Integrate: ln |u| + C.
Step 5. Replace u: ln |ex + 1| + C.
Step 6. Differentiate and Check: .
Example 7
Evaluate .
Step 1. Let u = 3x2.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 8
Evaluate .
Step 1. Let u = 2x.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: (52x)(2) ln 5/2 ln 5 = 52x.
Example 9
Evaluate .
Step 1. Let u = x4.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 10
Evaluate .
Step 1. Let u = cos π x.
Step 2. Differentiate: du = −π sin π x dx; .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6: Differentiate and Check: .
10.3 Techniques of Integration
Main Concepts: Integration by Parts, Integration by Partial Fractions
Integration by Parts
According to the product rule for differentiation . Integrating tells us that , and therefore . To integrate a product, careful identification of one factor as u and the other as allows the application of this rule for integration by parts. Choice of one factor to be u (and therefore the other to be dv) is simpler if you remember the mnemonic LIPET. Each letter in the acronym represents a type of function: Logarithmic, Inverse trigonometric, Polynomial, Exponential, and Trigonometric. As you consider integrating by parts, assign the factor that falls earlier in the LIPET list as u, and the other as dv.
Example 1
Step 1: Identify u = x and dv = e−xdx since x is a Polynomial, which comes before Exponential in LIPET.
Step 2: Differentiate du = dx and integrate v = −e−x.
Step 3:
Example 2
Step 1: Identify u = x and dv = sin 4x dx since x is a Polynomial, which comes before Trigonometric in LIPET.
Step 2: Differentiate du = dx and integrate .
Step 3:
Integration by Partial Fractions
A rational function with a factorable denominator can be integrated by decomposing the integrand into a sum of simpler fractions. Each linear factor of the denominator becomes the denominator of one of the partial fractions.
Example 1
Step 1: Factor the denominator:
Step 2: Let A and B represent the numerators of the partial fractions .
Step 3: The algorithm for adding fractions tells us that A(x − 1) + B(x + 4) = 1, so Ax + Bx = 0 and − A + 4B = 1. Solving gives us A = −0.2 and B = 0.2.
Step 4: ln
Example 2
Step 1: Use long division to rewrite .
Step 2: Factor the denominator:
Step 3: Let A, B, and C represent the numerators of the partial fractions.
Step 4: 2x2 + x + 1 = A(x + 1)(x − 1) + Bx(x − 1) + Cx(x + 1), therefore, Ax2 + Bx2 + Cx2 = 2x2, Cx − Bx = x, and −A = 1. Solving gives A = −1, B = 1, and C = 2.
Step 5:
10.4 Rapid Review
10.4 U-substitution Trig Functionsap Calculus Answers
1. Evaluate .
Answer: Rewrite as .
2. Evaluate .
Answer: Rewrite as .
3. Evaluate .
Answer: Rewrite as .
Thus, .
4. Evaluate .
Answer: −cos x + C.
5. Evaluate .
Answer: Let u = 2x and obtain .
6. Evaluate .
Answer: Let u = ln x; and obtain .
7. Evaluate .
Answer: Let u = x2; and obtain .
8.
Answer: Let u = x, du = dx, dv = cos xdx, and v = sin x, then .
9.
Answer:
10.5 Practice Problems
Evaluate the following integrals in problems 1 to 20. No calculators are allowed. (However, you may use calculators to check your results.)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17. If and the point (0, 6) is on the graph of y, find y.
18.
19.
20. If f(x) is the antiderivative of and f(1) = 5, find f(e).
21.
22.
23.
24.
25.
10.6 Cumulative Review Problems
(Calculator) indicates that calculators are permitted.
26. The graph of the velocity function of a moving particle for 0 ≤ t ≤ 10 is shown in Figure 10.6-1.
Figure 10.6-1
(a) At what value of t is the speed of the particle the greatest?
(b) At what time is the particle moving to the right?
27. Air is pumped into a spherical balloon, whose maximum radius is 10 meters. For what value of r is the rate of increase of the volume a hundred times that of the radius?
28. Evaluate .
29. (Calculator) The function f is continuous and differentiable on (0, 2) with f″(x) > 0 for all x in the interval (0, 2). Some of the points on the graph are shown below.
Which of the following is the best approximation for f′(1)?
(a) f′(1) < 2
(b) 0.5 < f′(1) < 1
(c) 1.5 < f′(1) < 2.5
(d) 2.5 < f′(1) < 3.5
(e) f′(1) > 2
30. The graph of the function f″ on the interval [1, 8] is shown in Figure 10.6-2. At what value(s) of t on the open interval (1, 8), if any, does the graph of the function f′:
Figure 10.6-2
(a) have a point of inflection?
(b) have a relative maximum or minimum?
10.4 U-substitution Trig Functionsap Calculus Pdf
(c) become concave upward?
31. Evaluate .
32. If the position of an object is given by x = 4 sin(πt), y = t2 − 3t + 1, find the position of the object at t = 2.
33. Find the slope of the tangent line to the curve r = 3 cos θ when .
10.7 Solutions to Practice Problems
1.
2. Rewrite:
3. Let u = x4 − 10du = 4x3dx or .
Rewrite
.
4. Let u = x2 + 1 ⇒ (u − 1) = x2 and du = 2x dx or .
5. Let u = x − 1; du = dx and (u + 1) = x.
6. Let ; or 2du = dx.
7. Let u = x2; du = 2x dx or .
8. Let u = cos x; du = −sin x dx or −du = sin x dx.
9.
Let u = x + 1; du = dx.
10.
11. Rewrite: .
Let u = 6x; du = 6 dx or .
12. Let u = ln x; .
13. Since ex and ln x are inverse functions:
14.
Let u = 3x; du = 3dx;
Let v = −x; dv = −dx;
Thus .
Note: C1 and C2 are arbitrary constants, and thus C1 + C2 = C.
15. Rewrite:
16. Let u = 1 + x3/2; or .
Rewrite:
17. Since , then y = .
The point (0, 6) is on the graph of y. Thus, 6 = e0 + 2(0) + C ⇒ 6 = 1 + C or C = 5. Therefore, y = ex + 2x + 5.
U-Substitution and Algebraic Functions
Another Form of the Integral of a Composite Function
If f is a differentiable function, then
Making a U-Substitution
Let u = f(x); then du = f′(x)dx.
Example 1
Evaluate .
Step 1. Let u = x + 1; then x = u − 1.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check:
Example 2
Evaluate .
Step 1. Let u = x − 2; then x = u + 2.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace: .
Step 6. Differentiate and Check:
Example 3
Evaluate .
Step 1. Let u = 2x − 5.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 4
Evaluate .
Step 1. Let u = x3 − 8.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
U-Substitution and Trigonometric Functions
Example 1
Evaluate .
Step 1. Let u = 4x.
Step 2. Differentiate: du = 4 dx or .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 2
Evaluate .
Step 1. Let u = tan x.
Step 2. Differentiate: du = sec2x dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: 2(tan x)3/2 + C or 2 tan3/2x + C.
Step 6. Differentiate and Check: .
Example 3
Evaluate .
Step 1. Let u = x3.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
• Remember that the area of a semi-circle is . Do not forget the . If the cross sections of a solid are semi-circles, the integral for the volume of the solid will involve which is .
U-Substitution and Inverse Trigonometric Functions
Example 1
Evaluate .
Step 1. Let u = 2x.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check:
Example 2
Evaluate .
Step 1. Rewrite:
.
Let u = x + 1.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check:
• If the problem gives you that the diameter of a sphere is 6 and you are using formulas such as or s = 4π r2, do not forget that r = 3.
U-Substitution and Logarithmic and Exponential Functions
Example 1
Evaluate .
Step 1. Let u = x4 − 1.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 2
Evaluate .
Step 1. Let u = cos x + 1.
Step 2. Differentiate: du = −sin xdx ⇒ − du = sin xdx.
Step 3. Rewrite: .
Step 4. Integrate: − ln |u| + C.
Step 5. Replace u: − ln |cos x + 1| + C.
Step 6. Differentiate and Check: .
Example 3
Evaluate .
Step 1. Rewrite ; by dividing (x2 + 3) by (x − 1).
Let u = x − 1.
Step 2. Differentiate: du = dx.
Step 3. Rewrite: .
Step 4. Integrate: 4 ln |u| + C.
Step 5. Replace u: 4 ln |x − 1| + C.
.
Step 6. Differentiate and Check:
.
Example 4
Evaluate .
Step 1. Let u = ln x.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 5
Evaluate .
Step 1. Let u = 2x − 5.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 6
Evaluate .
Step 1. Let u = ex + 1.
Step 2. Differentiate: du = exdx.
Step 3. Rewrite: .
Step 4. Integrate: ln |u| + C.
Step 5. Replace u: ln |ex + 1| + C.
Step 6. Differentiate and Check: .
Example 7
Evaluate .
Step 1. Let u = 3x2.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 8
Evaluate .
Step 1. Let u = 2x.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: (52x)(2) ln 5/2 ln 5 = 52x.
Example 9
Evaluate .
Step 1. Let u = x4.
Step 2. Differentiate: .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6. Differentiate and Check: .
Example 10
Evaluate .
Step 1. Let u = cos π x.
Step 2. Differentiate: du = −π sin π x dx; .
Step 3. Rewrite: .
Step 4. Integrate: .
Step 5. Replace u: .
Step 6: Differentiate and Check: .
10.3 Techniques of Integration
Main Concepts: Integration by Parts, Integration by Partial Fractions
Integration by Parts
According to the product rule for differentiation . Integrating tells us that , and therefore . To integrate a product, careful identification of one factor as u and the other as allows the application of this rule for integration by parts. Choice of one factor to be u (and therefore the other to be dv) is simpler if you remember the mnemonic LIPET. Each letter in the acronym represents a type of function: Logarithmic, Inverse trigonometric, Polynomial, Exponential, and Trigonometric. As you consider integrating by parts, assign the factor that falls earlier in the LIPET list as u, and the other as dv.
Example 1
Step 1: Identify u = x and dv = e−xdx since x is a Polynomial, which comes before Exponential in LIPET.
Step 2: Differentiate du = dx and integrate v = −e−x.
Step 3:
Example 2
Step 1: Identify u = x and dv = sin 4x dx since x is a Polynomial, which comes before Trigonometric in LIPET.
Step 2: Differentiate du = dx and integrate .
Step 3:
Integration by Partial Fractions
A rational function with a factorable denominator can be integrated by decomposing the integrand into a sum of simpler fractions. Each linear factor of the denominator becomes the denominator of one of the partial fractions.
Example 1
Step 1: Factor the denominator:
Step 2: Let A and B represent the numerators of the partial fractions .
Step 3: The algorithm for adding fractions tells us that A(x − 1) + B(x + 4) = 1, so Ax + Bx = 0 and − A + 4B = 1. Solving gives us A = −0.2 and B = 0.2.
Step 4: ln
Example 2
Step 1: Use long division to rewrite .
Step 2: Factor the denominator:
Step 3: Let A, B, and C represent the numerators of the partial fractions.
Step 4: 2x2 + x + 1 = A(x + 1)(x − 1) + Bx(x − 1) + Cx(x + 1), therefore, Ax2 + Bx2 + Cx2 = 2x2, Cx − Bx = x, and −A = 1. Solving gives A = −1, B = 1, and C = 2.
Step 5:
10.4 Rapid Review
10.4 U-substitution Trig Functionsap Calculus Answers
1. Evaluate .
Answer: Rewrite as .
2. Evaluate .
Answer: Rewrite as .
3. Evaluate .
Answer: Rewrite as .
Thus, .
4. Evaluate .
Answer: −cos x + C.
5. Evaluate .
Answer: Let u = 2x and obtain .
6. Evaluate .
Answer: Let u = ln x; and obtain .
7. Evaluate .
Answer: Let u = x2; and obtain .
8.
Answer: Let u = x, du = dx, dv = cos xdx, and v = sin x, then .
9.
Answer:
10.5 Practice Problems
Evaluate the following integrals in problems 1 to 20. No calculators are allowed. (However, you may use calculators to check your results.)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17. If and the point (0, 6) is on the graph of y, find y.
18.
19.
20. If f(x) is the antiderivative of and f(1) = 5, find f(e).
21.
22.
23.
24.
25.
10.6 Cumulative Review Problems
(Calculator) indicates that calculators are permitted.
26. The graph of the velocity function of a moving particle for 0 ≤ t ≤ 10 is shown in Figure 10.6-1.
Figure 10.6-1
(a) At what value of t is the speed of the particle the greatest?
(b) At what time is the particle moving to the right?
27. Air is pumped into a spherical balloon, whose maximum radius is 10 meters. For what value of r is the rate of increase of the volume a hundred times that of the radius?
28. Evaluate .
29. (Calculator) The function f is continuous and differentiable on (0, 2) with f″(x) > 0 for all x in the interval (0, 2). Some of the points on the graph are shown below.
Which of the following is the best approximation for f′(1)?
(a) f′(1) < 2
(b) 0.5 < f′(1) < 1
(c) 1.5 < f′(1) < 2.5
(d) 2.5 < f′(1) < 3.5
(e) f′(1) > 2
30. The graph of the function f″ on the interval [1, 8] is shown in Figure 10.6-2. At what value(s) of t on the open interval (1, 8), if any, does the graph of the function f′:
Figure 10.6-2
(a) have a point of inflection?
(b) have a relative maximum or minimum?
10.4 U-substitution Trig Functionsap Calculus Pdf
(c) become concave upward?
31. Evaluate .
32. If the position of an object is given by x = 4 sin(πt), y = t2 − 3t + 1, find the position of the object at t = 2.
33. Find the slope of the tangent line to the curve r = 3 cos θ when .
10.7 Solutions to Practice Problems
1.
2. Rewrite:
3. Let u = x4 − 10du = 4x3dx or .
Rewrite
.
4. Let u = x2 + 1 ⇒ (u − 1) = x2 and du = 2x dx or .
5. Let u = x − 1; du = dx and (u + 1) = x.
6. Let ; or 2du = dx.
7. Let u = x2; du = 2x dx or .
8. Let u = cos x; du = −sin x dx or −du = sin x dx.
9.
Let u = x + 1; du = dx.
10.
11. Rewrite: .
Let u = 6x; du = 6 dx or .
12. Let u = ln x; .
13. Since ex and ln x are inverse functions:
14.
Let u = 3x; du = 3dx;
Let v = −x; dv = −dx;
Thus .
Note: C1 and C2 are arbitrary constants, and thus C1 + C2 = C.
15. Rewrite:
16. Let u = 1 + x3/2; or .
Rewrite:
17. Since , then y = .
The point (0, 6) is on the graph of y. Thus, 6 = e0 + 2(0) + C ⇒ 6 = 1 + C or C = 5. Therefore, y = ex + 2x + 5.
18. Let u = ex; du = ex dx.
Rewrite: −3
19. Let u = ex + e−x; du = (ex − e−x) dx.
20. Since f(x) is the antiderivative of , .
Given f(1) = 5; thus, ln (1) + C = 5 ⇒ 0 + C = 5 or C = 5.
Thus, f(x) = ln |x| + 5 and f(e) = ln (e) + 5 = 1 + 5 = 6.
21. Integrate by parts. Let u = x2, du = 2x dx, , and . Then . Use parts again with u = x, du = dx, dv= (1 − x)3/2dx, and so that . Integrate for and simplify to .
22. For , use integration by parts with u = 3x2, du = 6x dx, dv = sin x dx, and v = −cos x.
10.4 U-substitution Trig Functionsap Calculus Problems
23. Factor the denominator so that . Use a partial fraction decomposition, , which implies Ax + A + Bx − 4B = x. Solve A + B = 1 and A − 4B to find and . Integrate .
24. Factor and use partial fractions. If , Ax + A + Bx = 1 and A = −B = 1.
25. Begin with integration by parts, using u = ln x, , , and .
Then
.
Use a partial fraction decomposition . Solve to find . Then
10.8 Solutions to Cumulative Review Problems
26. (a) At t = 4, speed is 5 which is the greatest on 0 ≤ t ≤ 10.
(b) The particle is moving to the right when 6 < t< 10.
27.
If , then .
Since r ≥ 0, meters.
28. Let u = ln x; .
29. Label given points as A, B, C, D, and E. Since f″(x) > 0 ⇒ f is concave upward for all x in the interval [0, 2]. Thus, and . Therefore, 1.5 < f′(1) < 2.5, choice (c). (See Figure 10.8-1.)
Figure 10.8-1
30. (a) f″ is decreasing on [1, 6) ⇒ f″′< 0 ⇒ f′ is concave downward on [1, 6) and f″ is increasing on (6, 8] ⇒ f′ is concave upward on (6, 8]. Thus, at x = 6, f′ has a change of concavity. Since f″ exists at x = 6 (which implies there is a tangent to the curve of f′ at x = 6), f′ has a point of inflection at x = 6.
10.4 U-substitution Trig Functionsap Calculus Worksheet
(b) f″ > 0 on [1, 4] ⇒ f′ is increasing and f″ < 0 on (4, 8] ⇒ f′ is decreasing. Thus at x = 4, f′ has a relative maximum at x = 4. There is no relative minimum.
(c) f″ is increasing on [6, 8] ⇒ f′ > 0 ⇒ f′ is concave upward on [6, 8].
31.
32. At t = 2, x = 4 sin(2π) = 0, and , so the position of the object at t = 2 is (0, −1).
33. To find the slope of the tangent line to the curve r = 3 cos θ when , begin with x = r cos θ and y = r sinθ, and find and so . y = 3 cos θ sinθ so . Then the slope of the tangent line is . . Evaluate at to get . The slope of the tangent line is zero, indicating that the tangent is horizontal.
