Integrating e^x and 1/x

6 Key excerpts on "Integrating e^x and 1/x"

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  eBook - PDF

  Applied Calculus

  • Geoffrey Berresford, Andrew Rockett(Authors)
  • 2015(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  c. To find distance, we speed. d. To find acceleration, we speed. 69. For each of the following casual, everyday sentences, fill in the blank with one of the words differentiate or integrate , whichever is more appropriate. a. To find marginal cost, we cost. b. To find profit, we marginal profit. c. To find how fast a stock price is rising, we price. d. To find a population, we how fast it is growing. Introduction In the previous section we defined integration as the reverse of differentiation, and we introduced several integration formulas. In this section we develop integration formulas involving logarithmic and exponential functions. One of these formulas will answer a question that we could not answer earlier—namely, how to integrate x 2 1 , the only power not covered by the Power Rule. The Integral # e ax dx On page 279 we saw that to differentiate e ax , we multiply by a to get ae ax . Therefore, to integrate e ax , the reverse process, we must divide by a. That is, for any a Þ 0 : 5.2 Integration Using Logarithmic and Exponential Functions Integrating an Exponential Function # e ax dx 5 1 a e ax 1 C Brief Example # e 2 x dx 5 1 2 e 2 x 1 C Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 320 Chapter 5 Integration and Its Applications In words: The integral of e to a constant times x is 1 over the constant times the original function (plus C ). As always, we may check the answer by differentiation. d dx a 1 2 e 2 x 1 C b 5 1 2 # 2 # e 2 x 5 e 2 x Using d dx e ax 5 ae ax and canceling the 2’s The result is the integrand e 2x , so the integration is correct.

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  Introduction to Integral Calculus

  Systematic Studies with Engineering Applications for Beginners

  • Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, A. K. Ghosh(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  All the properties of exponential function studied there can be established using the definition of the logarithmic function, discussed in this chapter. It will be seen that these properties are consistent with the properties of exponents learnt in algebra. 196 THE INTEGRAL FUNCTION Ð x 1 1 t dt, (x > 0) IDENTIFIED AS ln x OR log e x 7a Methods for Evaluating Definite Integrals 7a.1 INTRODUCTION In Chapter 5, we have introduced the following concepts/ideas: . the concept of area, . the meaning of the definite integral as an area, . the idea of the definite integral as the limit of a sum, . the concept of Riemann sums and the analytical definition of definite integral as the limit of Riemann sums. . the symbol Ð b a f ðxÞdx for the definite integral of a (continuous) function f(x) defined on a closed interval [a, b], . the statement of the integrability theorem, and . the statement of the second fundamental theorem of Calculus, which links the definite integral Ð b a f ðxÞdx with the antiderivative Ð f ðxÞdx. Historically, methods of computing areas of certain regions were developed by the ancient Greeks. Such an area was called the integral. The symbol “ Ð ” [with positive numbers a and b (a < b)] was used to indicate the measure of the area in question. Thus, the term “integral” and the symbol “ Ð ” were in use prior to the discovery of differential Calculus. The concept of derivatives was discovered in the seventeenth century and the methods for finding the derivatives of various functions, were developed then. Simultaneously, mathema- ticians developed the concept of the antiderivative of a function and the methods for finding antiderivatives. The methods for computing derivatives of functions, together with those for computing antiderivatives constituted the subject of Calculus. Of course, a number of mathematicians have contributed through the centuries, towards the development of Calculus.

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  Essential Mathematics for Economics and Business

  • Teresa Bradley(Author)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  The use of |x | will be important in the section on definite integration. In this section, on indefinite integration, we simply write  1 x dx = ln(x ) + c The integral of a constant term The integral of a constant term (K) is deduced as follows:  K dx =  K x 0 dx since x 0 = 1 Now apply the power rule. When n = 0 → n + 1 = 0 + 1 = 1, therefore  K x 0 dx = K  x 0 dx = K x 1 1 + c = K x + c  K dx = K x + c (8.5) In other words, the integral of a constant term is the constant term multiplied by the independent variable. Working rules for integration The following working rules (similar to those used in differentiation) apply when integrating sums and differences of several functions.  As in differentiation, the sum or difference of several power functions is integrated by integrating each function separately:  {f(x ) + g(x )}dx =  f(x )dx +  g(x )dx (8.6)  As in differentiation, the integral of a constant multiplied by a variable term is the constant multiplied by the integral of the variable term:  K f(x )dx = K  f(x )dx (8.7) [ 433 ] I NTEGRATION AND A PPLICATIONS WORKED EXAMPLE 8.2 INTEGRATING SUMS AND DIFFERENCES, CONSTANT MULTIPLIED BY VARIABLE TERM Integrate (a) x 2 + 1 x 2 and (b) 5x − 3x 2 . SOLUTION (a) Step 1: Simplify, x 2 + 1 x 2 = x 2 + x −2 Step 2: State n, hence n + 1 for each term for x 2 we have n = 2 → n + 1 = 3 for x −2 we have n = −2 → n + 1 = −2 + 1 = −1 Step 3: Apply the power rule to obtain  (x 2 + x −2 )dx = x 3 3 +  x −1 −1  = x 3 3 − x −1 1 = x 3 3 − 1 x + c (b) The steps are worked through very briefly (explain each step to yourself):  (5x − 3x 2 )dx =  5x dx −  3x 2 dx = 5  x dx − 3  x 2 dx = 5 x 2 2 − 3 x 3 3 = 2.5x 2 − x 3 + c Note 1: You can easily check that your integration is correct. According to equation (8.2), d  (function)  = function. So differentiate your answer and you should end up with the function you originally set out to integrate.

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  350 Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions LOGARITHMIC DIFFERENTIATION We now consider a technique called logarithmic differentiation that is useful for differen- tiating functions that are composed of products, quotients, and powers. Example 5 The derivative of y = x 2 3 √ 7x − 14 (1 + x 2 ) 4 (7) is messy to calculate directly. However, if we first take the natural logarithm of both sides and then use its properties, we can write ln y = 2 ln x + 1 3 ln(7x − 14) − 4ln(1 + x 2 ) Differentiating both sides with respect to x yields 1 y dy dx = 2 x + 7 / 3 7x − 14 − 8x 1 + x 2 Thus, on solving for dy /dx and using (7) we obtain dy dx = x 2 3 √ 7x − 14 (1 + x 2 ) 4  2 x + 1 3x − 6 − 8x 1 + x 2  REMARK Since ln y is only defined for y > 0, the computations in Example 5 are only valid for x > 2 (verify). However, using the fact that the expression for the derivative of ln y is the same as that for ln |y|, it can be shown that the formula obtained for dy / dx is valid for x < 2 as well as x > 2 (Exercise 75). In general, whenever a derivative dy / dx is obtained by logarithmic differentiation, the resulting derivative formula will be valid for all values of x for which y = 0. It may be valid at those points as well, but it is not guaranteed. INTEGRALS INVOLVING ln x Formula (2) states that the function ln x is an antiderivative of 1 / x on the interval (0, +∞), whereas Formula (6) states that the function ln |x| is an antiderivative of 1 / x on each of the intervals (−∞, 0) and (0, +∞). Thus we have the companion integration formula to (6),  1 u du = ln |u| + C (8) with the implicit understanding that this formula is applicable only across an interval that does not contain 0. Example 6 Applying Formula (8),  e 1 1 x dx = ln |x|] e 1 = ln |e| − ln |1| = 1 − 0 = 1  −1 −e 1 x dx = ln |x|] −1 −e = ln | −1| − ln | − e| = 0 − 1 = −1 Example 7 Evaluate  3x 2 x 3 + 5 dx.

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  eBook - PDF

  Calculus, Volume 1

  • Tom M. Apostol(Author)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  Another way is to define a x by the formula a x = e x log a . (6.34) The second method is preferable because, first of all, it is meaningful for all positive a (including a = 1) and, secondly, it makes it easy to prove the following properties of exponentials: log a x = x log a. (ab) x = a x b x . a x a y = a x+y . (a x ) y = (a y ) x = a xy . If a ≠ 1, then y = a x if and only if x = log a y. The proofs of these properties are left as exercises for the reader. Just as the graph of the exponential function was obtained from that of the logarithm by reflec- tion through the line y = x, so the graph of y = a x can be obtained from that of y = log a x by reflection through the same line; examples are shown in Figure 6.7. The curves in Figures 6.7 were obtained by reflection of those in Figures 6.3. The graph corresponding to a = 1 is, of course, the horizontal line y = 1. 6.16 Differentiation and integration formulas involving exponentials One of the most remarkable properties of the exponential function is the formula E ′ (x) = E(x), (6.35) which tells us that this function is its own derivative. If we use this along with the chain rule, we can obtain differentiation formulas for exponential functions with any positive base a. Suppose f (x) = a x for x > 0. By the definition of a x , we may write f (x) = e x log a = E(x log a); 246 The logarithm, the exponential, and the inverse trigonometric functions hence, by the chain rule, we find f ′ (x) = E ′ (x log a) . log a = E(x log a) . log a = a x log a. (6.36) In other words, differentiation of a x simply multiplies a x by the constant factor log a, this factor being 1 when a = e. y x 0 1 a = e a > e 1 < a < e (a) a > 1 y x 0 1 (b) 0 < a < 1 0 < a < a = 1 e 1 e < a < 1 1 e Figure 6.7 The graph of y = a x for various values of a. Of course, these differentiation formulas automatically lead to corresponding integration for- mulas.

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  Precalculus

  Functions and Graphs

  • Earl Swokowski, Jeffery Cole(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The natural exponential function is one of the most useful functions in advanced mathematics and applications. Since 2 , e , 3 , the graph of y 5 e x lies between the graphs of y 5 2 x and y 5 3 x , as shown in Figure 1. Scientific and graphing calculators have an e x key for approximating values of the nat- ural exponential function. FIGURE 1 y x y H11005 2 x y H11005 3 x y H11005 e x Continuously Compounded Interest The compound interest formula is A 5 P S 1 1 r n D nt . If we let 1 y k 5 r y n , then k 5 n y r , n 5 kr , and nt 5 krt , and we may rewrite the formula as A 5 P S 1 1 1 k D krt 5 P 3 S 1 1 1 k D k 4 rt . For continuously compounded interest we let n (the number of interest periods per year) increase without bound, denoted by n → ` or, equivalently, by k → ` . Using the fact that f 1 1 s 1 y k dg k → e as k → ` , we see that P 3 S 1 1 1 k D k 4 rt → P f e g rt 5 Pe r t as k → ` . This result gives us the following formula. The e x key can be accessed by pressing 2nd LN . Definition of the Natural Exponential Function The natural exponential function f is defined by f s x d 5 e x for every real number x . 276 CHAPTER 4 Inverse, Exponential, and Logarithmic Functions Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The next two examples illustrate the use of this formula.

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