Review Topics
C ertain topics in algebra, geometry, analytical geometry, and trigonom- etry are essential in preparing to study calculus Some of them are briefly reviewed in the following sections.
The set of real numbers (R) is crucial in calculus, encompassing both rational and irrational numbers Interval notation offers a concise way to represent intervals of real numbers, eliminating the need for inequality symbols or set-builder notation.
The following lists some common intervals of real numbers and their equivalent expressions, using set-builder notation:
Note that an infinite end point ^!3his never expressed with a bracket in interval notation because neither + 3nor - 3represents a real number value.
The absolute value is a crucial concept in calculus, representing the distance of a number x from zero on a real number line, regardless of direction This definition ensures that the absolute value is always nonnegative, meaning x is greater than or equal to zero.
A common algebraic definition of absolute value is often stated in three parts, as follows:
In calculus, the equation x = x² is often associated with the principal square root of x², which indicates that the absolute value of a number is always nonnegative.
A function is a collection of ordered pairs (x, y), where each first element x is associated with exactly one second element y The first elements form the function's domain, while the second elements constitute its range In this context, the domain variable is known as the independent variable, and the range variable is referred to as the dependent variable The notation f(x) is commonly used to represent the value of the function f for a specific x, and it is pronounced as "f of x" or "f at x."
A graph represents a function if any vertical line intersects it at most once, ensuring that each x value corresponds to only one y value If a vertical line intersects the graph at multiple points, it indicates that a single x value is linked to multiple y values, violating the definition of a function Understanding these principles is essential, as many key concepts and theorems in calculus are directly related to functions.
Example 1-1: The following are some examples of equations that are functions.
Example 1-2: The following are some equations that are not functions; each has an example to illustrate why it is not a function.
(c)x= -5 If x= -5, then y can be any real number.
A linear equation is defined as an equation that can be represented in the standard form ax + by = c, where at least one of the coefficients a or b is non-zero Even if a linear equation is not initially in this format, it can be rearranged algebraically The slope of a line, often represented by the letter m, indicates the direction of the line, whether it rises, falls, or remains horizontal or vertical.
Chapter 1: Review Topics 5 m=sunsunriserise horizontal horizontal changechange vertical vertical changechange value value changechange value value changechange x
A vertical line has a constant x value and a horizontal change of zero, resulting in an undefined or nonexistent slope In contrast, nonvertical lines possess a numerical slope: a positive slope indicates an upward slant to the right, a negative slope indicates a downward slant to the right, and a slope of zero signifies a horizontal line.
Example 1-3: Find the slope of the line passing through (–5, 4) and (–1, –3). m x x y y
= -7 The line, then, has a slope of –7/4.
Certain representations of linear equations are designated with specific names based on their structure Although these forms may seem distinct, they can be algebraically transformed to demonstrate their equivalence.
Any nonvertical lines are parallel if they have the same slopes, and con- versely lines with equal slopes are parallel If the slopes of two lines L 1 and
L 2 are m 1 and m 2, respectively, then L 1is parallel to L 2if and only if m 1 = m 2
Two lines that are neither vertical nor horizontal are considered perpendicular if the product of their slopes equals –1 Specifically, if the slopes of lines L1 and L2 are represented as m1 and m2, then L1 is perpendicular to L2 if and only if m1 multiplied by m2 equals –1 Additionally, it's important to remember that any two vertical lines are parallel, while a vertical line is always perpendicular to a horizontal line.
The standard form of a linear equation is expressed as ax + by = c, where a and b cannot both be zero When b equals zero, the equation simplifies to x = constant, indicating a vertical line Conversely, if a equals zero, the equation becomes y = constant, representing a horizontal line.
Example 1-4: The following are some examples of linear equations expressed in general form:
The point-slope form of a linear equation is y-y 1=m x( -x 1)when the line passes through the point (x 1,y 1) and has a slope m.
Example 1-5: Find an equation of the line through the point (3,4) with slope –2/3.
The slope-intercept form of a linear equation is y = mx + b when the line has y-intercept (0,b) and slope m.
Example 1-6: Find an equation of the line that has a slope 4/3 and crosses y-axis at –5. y=mxmx+b
The intercept form of a linear equation is x/a + y/b = 1 when the line has x-intercept (a,0) and y-intercept (0,b).
Example 1-7: Find an equation of the line that crosses the x-axis at –2 and the y-axis at 3. ax b y 1 + x y
In trigonometry, angle measure is expressed in one of two units: degrees or radians The relationship between these measures may be expressed as follows: 180180 % =rradians
To convert degrees to radians, the formula \( \text{radians} = \text{degrees} \times \frac{\pi}{180} \) is utilized Conversely, to change radians to degrees, the equation \( \text{degrees} = \text{radians} \times \frac{180}{\pi} \) is applied These relationships facilitate the conversion between these two units of angular measurement effectively.
180rto obtain the degree measure.
The six fundamental trigonometric functions can be defined through a circle represented by the equation x² + y² = r² These functions relate to an angle in standard position, which has its vertex at the circle's center and its initial side aligned with the positive x-axis.
The trigonometric functions sine, cosine, tangent, cotangent, secant, and cosecant are defined as follows:
Figure 1-1 Defining the trigonometric functions. sin sin r y x y y
= i + tan tan coscos sin sin x
= = y i ii cot cot sinsin cos cos xy
Understanding the values of trigonometric functions at key angles such as 30°, 45°, 60°, 90°, and 180° (or in radians: π/6, π/4, π/3, π/2, and π) is crucial for calculus Additionally, familiarity with the graphs of the six trigonometric functions enhances comprehension Key trigonometric identities also play a significant role in calculus studies.
The article discusses the intricate relationships between sine, cosine, tangent, secant, and cotangent functions in trigonometry It emphasizes the importance of understanding how these functions interrelate and their applications in various mathematical contexts The repetition of these functions highlights their fundamental role in solving trigonometric equations and modeling periodic phenomena By mastering these concepts, one can enhance their mathematical proficiency and apply these principles effectively in real-world scenarios.
The relationship between the angles and sides of a triangle may be expressed using the Law of Sines or the Law of Cosines (see Figure 1-2).
Figure 1-2 Relations between sides and angles of a triangle.
: sin sin sinsin sinsin sin sin sinsin sinsin cos cos cos cos cos cos sin sin b aA b
Law ofof SinesSines or or
Law ofof CoCo eses or or or or
Table 1-1 Values of Sine, Cosine, and Tangent at
Radian Measure of x sin x cos x tan x
1 Which of the following equations is not a function? a 3x – 2y = 6 b y = sin 3x c y = x 2 d x 2 + y 2 = 16 e y= 3+x
2 Find an equation in general form of the line with slope –2/5 and y-intercept (0,3).
3 Find an equation in general form of the line passing through the origin and perpendicular to the line 5x – 3y = 6.
4 Find an equation in general form of the line passing through the points (3,–2) and (–1,0).
5 If θis an angle between π/2 and π, and sin θ= 3/5, what is cos θ? Answers: 1 d 2 2x + 5y = 15 3 3x + 5y = 0 4 x + 2y = –1 5 –4/5
❑ Determining when a function is continuous
The limit of a function is a fundamental concept in calculus, serving as the foundation for key ideas such as continuity, derivatives, and definite integrals Understanding limits is crucial for mastering these essential calculus principles.
Limits
❑ Finding derivatives of more complicated functions
The derivative of a function, a key application of limits in calculus, is crucial for solving a diverse range of problems Mastering the concept of derivatives is essential for effectively applying them in various mathematical scenarios.
The derivative of a function y = f (x) at a point (x,f (x)) is defined as
D if this limit exists The derivative is denoted by f'(x), read “f prime of x” or “f prime at x,” and f is said to be differentiable at x if this limit exists (see Figure 3-1).
Figure 3-1 The derivative of a function as the limit of rise over run. y x f(x + ∆x)
A function that is differentiable at a point x is guaranteed to be continuous at that point; however, the opposite is not always valid Specifically, a function can be continuous at a certain point while lacking a derivative at that same point For instance, the function f(x) = x^(1/3) is continuous across all real numbers, yet its derivative is undefined at x = 0.
Another example is the function f x( )= x+2, which is also continuous over its entire domain of real numbers but is not differentiable at x = –2
The relationship between continuity and differentiability can be summa- rized as follows: Differentiability implies continuity, but continuity does not imply differentiability.
Example 3-1: Find the derivative of f (x) = x 2 – 5 at the point (2,–1).
^ h h hence, the derivative of f (x) = x 2 – 5 at the point (2,–1) is 4.
The derivative of a function at a specific point represents the slope of the tangent line at that point It can be understood as the limit of the slopes of secant lines that connect a fixed point on the curve to other points that approach it If this limit exists, it is defined as the slope of the tangent line at the point (x, f(x)) on the graph of y = f(x).
The derivative can be understood as the instantaneous velocity of a particle's position along a line at a specific time t, represented by the function y = s(t) It is calculated as the limit of average velocities between a fixed time and times that approach it When this limit exists, it defines the instantaneous velocity at time t for the function y = s(t).
The derivative represents the instantaneous rate of change of a function at a specific point It can be understood as the limit of the average rates of change between a fixed point and nearby points on the curve, as those points approach the fixed point When this limit exists, it defines the instantaneous rate of change at the point (x, f(x)) on the graph of y = f(x).
Example 3-2: Find the instantaneous velocity of s t( ) t 2
^ h hence, the instantaneous velocity of s(t) = 1/(t + 2) at time t = 3 is –1/25. The negative velocity indicates that the particle is moving in the negative direction.
Various notations exist for representing the derivative of a function y = f(x), with f'(x) being the most widely used Other notations include y', dy/dx, df/dx, df(x)/dx, Dxf, and Dxf(x) Familiarity with these different forms is essential for solving specific problems in calculus.
Numerous differentiation rules can be established through the limit definition of the derivative, which are instrumental in calculating the derivatives of relevant functions To simplify the process and avoid relying on the formal definition for each derivative application, a selection of the most useful formulas is provided.
(1) If f (x) = c, where c is a constant, the f'(x) = 0.
(7) Power Rule: If f x( )=x n ,thenthenf xl( )=nxnx n - 1
Example 3-5: dx dx dy dy y xx
Example 3-9: Find the slope of the tangent line to the curve y = 12 – 3x 2 at the point (–1,9).
Because the slope of the tangent line to a curve is the derivative, you find that y' = –6x; hence, at (–1,9), y' = 6, and the tangent line has slope 6 at the point (–1,9).
The six trigonometric functions also have differentiation formulas that can be used in application problems of the derivative The rules are summa- rized as follows:
( )6 IfIf f x( )=csccscx,thenthenf x( )= -csccsc cotxcotx.
( )3 IfIf f x( )=tantanx,thenthenf x( )=secsec x. ( )2 IfIf f x( )=coscosx,thenthenf x( )= -sinsinx.
( ) ( ) , ( ) sin sin coscos cot cot csccsc sec sec secsec tantan f x x f x x f x x f x x f x x f x x x
The rules from (3) to (6) can be demonstrated using the quotient rule, applied to the function represented in sine and cosine terms, as shown in the example below.
Example 3-10: Use the definition of the tangent function and the quo- tient rule to prove if f (x) = tan x, than f '(x) = sec 2 x.
( ) ( ) tan tan cos cossinsin cos cos cos cos coscos sinsin sinsin cos cos cos cos sinsin cos cos sec sec f x x xx f x x x x x x x x x x x
( ) cot cot csccsc cot cot csccsc y x x x x x x x x
Example 3-12: FindFindf ifif f x( ) sinsinx coscosx.
( ) coscos sinsin cos cos sinsin f x x x f
Example 3-13: Find the slope of the tangent line to the curve y = sin x at the point (π/2,1)
The derivative of a curve represents the slope of its tangent line, and for the function y = sin x, we find that y' = cos x At the point (π/2, 1), the derivative evaluates to y' = cos(π/2) = 0, indicating that the slope of the tangent line is horizontal at this coordinate This geometric interpretation reveals that the tangent line at (π/2, 1) is flat on the graph of y = sin x.
The chain rule is a fundamental technique used to calculate the derivative of composite functions, where the number of functions in the composition dictates the number of differentiation steps required For instance, when dealing with a composite function f(x), the application of the chain rule allows for systematic differentiation based on its structure.
Note that because two functions, g and h, make up the composite func- tion f, you have to consider the derivatives g' and h' in differentiating f (x).
If a composite function r(x) is defined as
In the composition function r, which consists of three functions—m, n, and p—it is essential to consider their respective derivatives, m', n', and p', when differentiating r(x) A recommended technique for differentiating composite functions involves working from the "outside to the inside," establishing a sequence for each derivative that needs to be calculated.
Example 3-15: Find f'(x) if f (x) = tan (sec x).
( ) sec sec secsec secsec tantan sec sec tantan secsec secsec f x x x x x x x
Example 3-16: sinsin ( ).). dx dx dy dy y x
( ) ( ) sin sin coscos cos cos sinsin dx dx dy dy x x x x
Example 3-18: Find the slope of the tangent line to a curve y = (x 2 – 3) 5 at the point (–1, –32).
Because the slope of the tangent line to a curve is the derivative, you find that
- - l l which represents the slope of the tangent line at the point (–1,–32).
In mathematics, some equations involving x and y do not clearly define y as a function of x, making it challenging to express y in terms of x, even if such a function exists In these cases, it is understood that a function y = f(x) satisfies the given equation Implicit differentiation is a valuable technique that enables the calculation of the derivative of y with respect to x without the need to solve for y explicitly This method requires the application of the chain rule whenever differentiating y, based on the assumption that y can be represented as a function of x.
Example 3-19: y -xyxy10. dx dx dy dy x
Find Find ifif 2 3 Differentiating implicitly with respect to x, you find that
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Example 3-20: Find y' if y = sin x + cos y.
Differentiating implicitly with respect to x, you find that
( ) cos cos sinsin sin sin coscos sin sin coscos sin sin cos cos y x y y y y y x y y x y y x
Example 3-21: Find y' at (–1,1) if x 2 + 3xy +y 2 = –1.
Differentiating implicitly with respect to x, you find that
Example 3-22: Find the slope of the tangent line to the curve x 2 + y 2 = 25 at the point (3,–4).
Because the slope of the tangent line to a curve is the derivative, differen- tiate implicitly with respect to x, which yields x y y y y x y yx yx
=- l l l hence, at (3,–4), y' = –3/–4 = 3/4, and the tangent line has slope 3/4 at the point (3,–4)
The derivative of a function y = f(x) is another function, denoted as y' = f'(x) By taking the derivative of f'(x), we obtain the second derivative, represented as f''(x) or f²(x) This process can be extended to find higher-order derivatives, such as the third and fourth derivatives To simplify notation, the nth derivative of f(x) is commonly expressed as f^(n)(x) = y^(n) Chapter 4 explores applications of the second derivative in curve sketching and in analyzing distance, velocity, and acceleration.
Example 3-23: Find the first, second, and third derivatives of f (x) = 5x 4 – 3x 3 + 7x 2 – 9x + 2
= = - l m n Example 3-24: Find the first, second, and third derivatives of y = sin 2 x.
( ) sin sin coscos cos cos coscos sinsin sinsin cos cos sinsin cos cos sinsin sinsin coscos sin sin coscos sinsin coscos sin sin coscos y x x y x x x x x x y x x x x x x x x x x
Differentiation of Inverse Trigonometric Functions
Each of the six fundamental trigonometric functions possesses corresponding inverse functions, provided that suitable restrictions are applied to their domains The derivatives of all inverse trigonometric functions are well-defined and can be summarized effectively.
( ) sin sin arcsinarcsin cos cos arccosarccos tan tan arctanarctan cot cot cotcot sec sec secsec csc csc csccsc f x x x f x f x x f x x x f x f x x f x x x f x f x x f x x x f x f x x f x x x f x f x f x x x f x x x f x f x f x x x
Exponential functions and their corresponding inverse functions, called logarithmic functions, have the following differentiation formulas:
( ) loglog ln ln ln ln ln ln f x e f x e f x a a a f x a a f x x f x x f x x a a f x a x
Note that the exponential function f (x) = e x has the special property that its derivative is the function itself, f '(x) = e x = f (x).
( ) ln ln ln ln ln ln y x x y x
( ) sin sin coscos sin sin cos cos cot cot f x x x x x f x x
Example 3-31: loglog ( ).). dx dx dy dy y x x
( )()( ) ln ln ln ln dx dx dy dy x x x dx dx dy dy x x x
5 Find the slope of the tangent line to the curve x 2 + xy + y 2 = 4 at the point (–2,2).
Answers: 1 15x 2 + 2x sin x + x 2 cos x + (e x x – e x )/x 2 2 dy/dx = (2x 3 – 1)/(x 4 – 2x + 1) 1/2 3 e 5-x – tan x 4 2 3 3 5 1
The Derivative
❑ Using the derivative to understand the graph of a function
❑ Locating maximum and minimum values of a function
❑ Approximating quantities by using derivatives
The derivative of a function is essential in calculus, with applications including curve sketching, finding maximum and minimum values, and addressing problems related to distance, velocity, and acceleration Additionally, it is useful for solving related rate problems and for approximating function values.
The derivative of a function at a specific point represents the slope of the tangent line at that point The normal line, which is perpendicular to the tangent line, has a slope that is the negative reciprocal of the tangent's slope Therefore, the slope of the normal line to the graph of f(x) is calculated as –1/f'(x).
Example 4-1: Find the equation of the tangent line to the graph of ( ) f x = x 2 +3at the point (–1,2).
At the point (–1,2), f'(–1) = –1/2 and the equationof the line is
- = - - + Example 4-2: Find the equation of the normal line to the graph of ( ) f x = x 2 +3at the point (–1, 2).
From Example 4-1, you find that f'(–1) = –1/2 and the slope of the nor- mal line is –1/f'(–1) = 2; hence, the equationof the normal line at the point (–1,2) is
Critical points of a function are significant in derivative application problems, as they occur where the derivative is either zero or undefined A point (x, f(x)) is classified as a critical point if x is within the function's domain and satisfies either f'(x) = 0 or f'(x) does not exist Geometrically, at a critical point, the tangent line may be horizontal, vertical, or nonexistent, indicating important changes in the function's behavior.
Example 4-3: Find all critical points of f (x) = x 4 – 8x 2
Because f (x) is a polynomial function, its domain is all real numbers.
= - = - = - l l hence, the critical points of f (x) are (–2,–16), (0,0), and (2,–16).
Example 4-4: Find all critical points of f (x) = sin x + cos x on [0,2π]. The domain of f (x) is restricted to the closed interval [0,2π].
, cos cos sinsin cos cos sinsin cos cos sinsin sin sin coscos sin sin coscos f x x x f x x x x x x f f
= - r r r r r r r r l l c c m m hence, the critical points of f x are( )are( / ,r4 2)andand(5r/ ,4 - 2).).
An important application of critical points is in determining possible max- imum and minimum values of a function on certain intervals The Extreme Value Theorem guarantees both a maximum and minimum value for a function under certain conditions It states the following:
If a function f (x) is continuous on a closed interval [a,b], then f (x) has both a maximum and minimum value on [a,b].
Chapter 4: Applications of the Derivative 45
To apply the Extreme Value Theorem, first confirm that the function is continuous on a closed interval Next, identify all critical points within this interval, and evaluate the function at these points as well as at the interval's endpoints The highest value obtained will represent the maximum, while the lowest value will indicate the minimum of the function over the specified interval.
Example 4-5: Find the maximum and minimum values of f (x) = sin x + cos x on [0,2π]
The function is continuous on the interval [0, 2π], with critical points identified as ar/24k and a5r/4 - 2k At the endpoints, the function values are f(0) = 1 and f(2π) = 1 Consequently, the maximum value of the function occurs at x = π/4, yielding f(x) = 2, while the minimum value is found at x = 5π/4, resulting in f(x) = -2.
Note that for this example the maximum and minimum both occur at critical points of the function.
Example 4-6: Find the maximum and minimum values of f (x) = x 4 – 3x 3 – 1 on [–2,2].
The function is continuous on [–2,2], and its derivative is f'(x) = 4x 3 – 9x 2
The only critical point of the function occurs at x = 0, resulting in the point (0, –1), since x = 9/4 is outside the interval [–2, 2] Evaluating the function at the endpoints gives f(2) = –9 and f(–2) = 39, establishing the maximum value of 39 at x = –2 and the minimum value of –9 at x = 2 This highlights the significance of closed intervals in identifying critical points.
The Mean Value Theorem connects the slope of a tangent line to a curve with the slope of a secant line that intersects two points on the curve at the boundaries of a given interval.
If a function f (x) is continuous on a closed interval [a,b] and differentiable on an open interval (a,b), then at least one number c ∈(a,b) exists such that
Figure 4-1 The Mean Value Theorem.
The theorem states that there exists at least one point on the curve between the endpoints (a, f(a)) and (b, f(b)) where the slope of the tangent line is equal to the slope of the secant line In cases where f(a) equals f(b), the theorem ensures the presence of at least one critical point within the open interval (a, b) where the derivative f'(c) equals zero.
Example 4-7: Verify the conclusion of the Mean Value Theorem for f (x) = x 2 – 3x – 2 on [–2,3].
The function is continuous on [–2,3] and differentiable on (–2,3) The slope of the secant line through the endpoint values is
The slope of the tangent line is
Because 1/2 ∈[–2,3], the c value referred to in the conclusion of the Mean Value Theorem is c = 1/2 y x a c b
Chapter 4: Applications of the Derivative 47
The derivative of a function is a crucial tool for analyzing its behavior across different intervals in its domain When f'(x) > 0 throughout an interval I, the function is considered to be increasing in that interval Conversely, if f'(x) < 0 in an interval I, the function is deemed to be decreasing It's important to note that the derivative is either zero or undefined solely at the critical points of the function, indicating that it remains positive or negative at all other points within its domain.
To identify the intervals where a function is increasing or decreasing, start by locating the critical points within the domain Next, evaluate the intervals on either side of these critical points to assess the sign of the derivative If the derivative, f'(x), is greater than zero, the function is increasing in that interval; conversely, if f'(x) is less than zero, the function is decreasing This analysis aids in creating an accurate sketch of the function's graph.
Example 4-8: For f (x) = x 4 – 8x 2 determine all intervals where f is increas- ing or decreasing.
The function f(x) has a domain that includes all real numbers, with critical points identified at x = -2, 0, and 2 By analyzing the intervals surrounding these critical points using the derivative f'(x) = 4x³ - 16x, one can determine the behavior of the function across these values.
( )>)> f x f x f x f x l l l l hence, f is increasing on (–2,0) and (2,+ ∞) and decreasing on (–∞,–2) and (0,2).
Example 4-9: For f (x) = sin x + cos x on [0,2π], determine all intervals where f is increasing or decreasing.
The function f(x) is defined within the closed interval [0, 2π], with critical points identified at π/4 and 5π/4 By analyzing the derivative f'(x) = cos x – sin x, we can determine the behavior of the function in the intervals surrounding these critical points.
E hence, f is increasing on [0, π/4) and (5π/4,2π] and decreasing on (π/4,5π/4).
First Derivative Test for Local Extrema
A function exhibits a local extremum at a critical point when the derivative changes sign Specifically, if the derivative shifts from positive, indicating an increasing function, to negative, indicating a decreasing function, the function has a local maximum at that critical point.
When the derivative of a function transitions from negative to positive at a critical point, it indicates a local minimum This method for identifying local maxima and minima is known as the First Derivative Test for Local Extrema However, it is important to note that the derivative may not always change signs, making it crucial to evaluate each interval surrounding the critical point.
Example 4-10: If f (x) = x 4 – 8x 2 , determine all local extrema for the function.
Applications of the Derivative
❑ Understanding and computing basic indefinite integrals
❑ Using more advanced techniques of integration
❑ Understanding and computing definite integrals
Antidifferentiation, or integration, is a crucial operation in calculus alongside differentiation These two processes can be viewed as inverses of each other, and the derivative rules previously covered will aid in deriving corresponding antiderivative rules The connection between antiderivatives and definite integrals is explored later in the chapter through the Fundamental Theorem of Calculus.
An antiderivative F(x) of a function f(x) is defined by the condition that F'(x) = f(x) for all x in the domain of f It is important to note that antiderivatives are not unique; a function can have infinitely many antiderivatives For instance, F(x) = x³, G(x) = x³ + 5, and H(x) = x³ - 2 are all antiderivatives of f(x) = 3x², as their derivatives are equal to f(x) across the domain These functions differ only by a constant value, which has a derivative of zero Therefore, if F(x) and G(x) are both antiderivatives of f(x) over an interval, they can be expressed as F(x) = G(x) + C, where C is a constant This implies that the graphs of F(x) and G(x) are identical apart from their vertical displacement.
The notation for all antiderivatives of a function f(x) is represented by the indefinite integral symbol ∫, expressed as ∫ f(x) dx = F(x) + C In this expression, f(x) is known as the integrand, while C denotes the constant of integration The complete expression F(x) + C is referred to as the indefinite integral of the function.
F with respect to the independent variable x Using the previous example of F(x) = x 3 and f (x) = 3x 2 , you find that #3x dx 2 dx=x 3 +C
The indefinite integral of a function is sometimes called the general antiderivative of the function as well.
Example 5-1: Find the indefinite integral of f (x) = cos x.
. sin sin coscos cos cos sinsin
Because thethe derivativederivative ofof = isis = writewrite
Example 5-2: Find the general antiderivative of f (x) = –8.
Because thethe derivativederivative ofof 8 isis 8 writewrite
Numerous integration formulas can be directly obtained from their corresponding derivative formulas; however, some integration challenges demand additional effort These more complex problems often involve techniques such as substitution and change of variables, integration by parts, trigonometric integrals, and trigonometric substitutions.
Most of the following basic formulas directly follow the differentiation rules that were discussed in preceding chapters.
1 # kfkf x dx( )dx=k # f x dx( )dx
2 # 8 f x( )!g x dx( ) B dx= # f x dx( )dx! # g x dx( )dx
16 #secsecx dxdx=lnln secsecx+tantanx +C
17 #csccscx dxdx= -lnln csccscx+cotcotx +C
20 secsec x x a dx dx a1arcarc ax C
Using formula (4) from the preceding list, you find that x dxdx x5 C
# 1 Because 1/ x=x - 1 2 / , using formula (4) from the preceding list yields x dxdx x dxdx x C
Applying formulas (1), (2), (3), and (4), you find that
# + Using formula (13), you find that lnln x dx dx x C
Using formula (19) with a = 5, you find that arctan arctan x dx dx x C
Substitution and change of variables
Substitution and change of variables are effective integration techniques for evaluating indefinite integrals that do not conform to standard formulas This method, akin to the chain rule in differentiation, is particularly useful for composite functions Typically, the inner function is substituted with a single variable, commonly denoted as u It is essential that the derivative or a constant multiple of the derivative of this inner function is present as a factor in the integrand.
The substitution technique is utilized to transform the integration problem into a new variable format, enabling the application of fundamental integration formulas While this method may appear more labor-intensive at first, it ultimately simplifies the evaluation of the indefinite integral.
Note that for the final answer to make sense, it must be written in terms of the original variable of integration.
Because the inside function of the composition is x 3 + 1, substitute with u x du du x dxdx du du x dxdx
Example 5-9: Evaluate #sinsin x dx( )5 dx.
Because the inside function of the composition is 5x, substitute with hence, ( )
( ) sin sin sinsin cos cos cos cos u x du du dxdx du du dxdx x dxdx u dudu u C x C
Because the inside function of the composition is 9 – x 2 , substitute with hence, u x du du x dxdx du du x dxdx x x dxdx u dudu u dudu u C u C x C
Integration by parts is a valuable technique for evaluating indefinite integrals that do not conform to standard formulas This method is particularly useful when the integrand consists of a single transcendental function or a product of an algebraic function and a transcendental function The fundamental formula for integration by parts is expressed as ∫u dv = uv - ∫v du.
# where u and v are differential functions of the variable of integration
When applying integration by parts, start by selecting dv as the most complex component of the integrand that can be easily integrated to determine v The remaining portion of the integrand will serve as the u function, which you will differentiate to obtain du This method aims to simplify the integral into the form v du.
# , which is easier to evaluate than the original integral.
Example 5-11: Evaluate # xsecsec 2 x dxdx. hence, sec sec tan tan sec sec tantan tantan tan tan coscos tan tan coscos u x dvdv x dxdx du du dxdx v x x x dxdx x x x dxdx x x x C x x x C
Example 5-12: Evaluate # x 4 1nx dxdx. hence, ln ln ln ln lnln ln ln ln ln u x dvdv x dxdx du du x dxdx v x x x dxdx x x x x dxdx x x x dxdx x x x C
Example 5-13: Evaluate #arctanarctanx dxdx. hence,
( ) arctan arctan arctan arctan arctanarctan arctan arctan lnln u x dvdv dxdx du du x dxdx v x x dxdx x x x x dxdx x x C
Integrating powers of trigonometric functions often requires manipulation to align them with basic integration formulas Familiarity with fundamental trigonometric identities, as discussed in Chapter 1, is crucial for rewriting the integrand into a more manageable form Similar to the technique of integration by parts, the objective is to identify an integral that is simpler to evaluate than the original.
Example 5-14: Evaluate #coscos 3 xsinsin 4 dxdx
( ) cos cos sinsin coscos sinsin coscos sin sin sinsin coscos sin sin sinsin coscos sin sin coscos sinsin coscos sin sin sinsin x dxdx x x x dxdx x x x dxdx x x x dxdx x x dxdx x x dxdx x x C
Example 5-15: Evaluate #secsec x dx 6 dx
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Example 5-16: Evaluate #sinsin x dx 4 dx.
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If an integrand contains a radical expression of the form
, , a 2 -x 2 a 2 +x 2 oror x 2 -a 2 , a specific trigonometric substitution may be helpful in evaluating the indefinite integral Some general rules to fol- low are
1 If the integrand contains a 2 -x 2 let x = a sin θ dx = a cos θdθ and a 2 -x 2 =acoscosθ
2 If the integrand contains a 2 +x 2 let x = a tan θ dx = a sec 2 θdθ and a 2 +x 2 =asecsecθ
3 If the integrand contains x 2 -a 2 let x = a sec θ dx = a sec θ tanθdθ and x 2 -a 2 =atantanθ
Right triangles can be utilized in various scenarios to derive expressions for the six trigonometric functions essential in evaluating indefinite integrals.
Because the radical has the form a 2 -x 2
( ) sin sin sinsin cos cos cos cos x a dx dx d x let let and and FigureFigure
( sinsin )()( coscos ) cos cos sin sin csc csc cot cot x x dx dx d d d C x x C x x C
( ) tan tan tantan sec sec sec sec a x x a dx dx d x
Because thethe radicalradical hashas thethe formform let let and and FigureFigure
Figure 5-2 Diagram for Example 5-18. hence, sec sec sec sec sec sec sec sec tantan ln ln ln ln x dx dx d d
The indefinite integral plays a crucial role in solving problems related to distance, velocity, and acceleration as functions of time It is important to understand that the derivative of a distance function indicates instantaneous velocity, while the derivative of a velocity function reveals instantaneous acceleration at a specific moment Furthermore, recognizing the inverse relationship between derivatives and indefinite integrals is essential; the indefinite integral of the acceleration function yields the velocity function, and the indefinite integral of the velocity function produces the distance function.
When an object is in free fall, it experiences an acceleration of -32 ft/sec² due to gravity The negative sign indicates that the velocity is decreasing over time, reflecting a negative rate of change in velocity By recognizing that velocity is the indefinite integral of acceleration, we can further analyze the motion of the falling object.
Now, at t = 0, the initial velocity (v 0) is
= = - + hence, because the constant of integration for the velocity in this situation is equal to the initial velocity, write v(t) = –32t + v 0.
Because the distance is the indefinite integral of the velocity, you find that
Now, at t = 0, the initial distance (s 0) is
Chapter 5: Integration 73 hence, because the constant of integration for the distance in this situa- tion is equal to the initial distance, write s(t) = –16t 2 + v 0(t) + s 0.
Example 5-19: A ball is thrown downward from a height of 512 feet with a velocity of 64 feet per second How long will it take for the ball to reach the ground?
From the given conditions, you find that hence,
/ / sec sec sec sec a t v s v t t s t t t ft ft ft ft ft ft
The distance is zero when the ball reaches the ground or
- - + - + - - + - = - hence, the ball will reach the ground 4 seconds after it is thrown.
Example 5-20: In the previous example, what will the velocity of the ball be when it hits the ground?
Because v(t) = –32(t) – 64 and it takes 4 seconds for the ball to reach the ground, you find that
The ball will impact the ground at a velocity of -192 ft/sec, indicating a negative rate of change in distance over time This negative velocity signifies that as time progresses, the distance to the ground decreases.
Example 5-21: A missile is accelerating at a rate of 4t m/sec 2 from a posi- tion at rest in a silo 35 m below ground level How high above the ground will it be after 6 seconds?
From the given conditions, you find that a(t) = 4t m/sec 2 , v 0 = 0 m/sec because it begins at rest, and s0 = –35 m because the missile is below ground level; hence, and
After 6 seconds, you find that s( )6 2( )6 3 -3535m9109m hence, the missile will be 109 m above the ground after 6 seconds.