Chapter 7. Integrals and Transcendental Functions

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1 7. The Logarithm Define as an Integral Chapter 7. Integrals an Transcenental Functions 7.. The Logarithm Define as an Integral Note. In this section, we introuce the natural logarithm function using efinite integrals. However, to justify using the logarithm terminology we must show that the function we introuce satisfies the usual properties of logarithms. We will o so using our efinition an the calculus properties which it satisfies. Definition. For x > 0, efine the natural logarithm function as ln x = x t t. Note. It follows from the efinition that for x, ln x is the area uner the curve y = /t for t [, x]. All we can currently tell from the efinition, is that lnx < 0 for x (0, ), ln = 0, an lnx > 0 for x (, ). We also see that ln x is an INCreasing function of x.

2 7. The Logarithm Define as an Integral 2 Figure 7. page 48 Definition. The number e is that number in the omain of the natural logarithm satisfying ln(e) =. Numerically, e Note. The number e is an example of a transcenental number (as oppose to an algebraic number). The number π is also transcenental. The text mentions this on page 422. The six trigonometric functions, the logarithm functions, an exponential functions (to be efine soon) are transcenental (thus the title of this chapter).

3 7. The Logarithm Define as an Integral 3 Theorem. For x > 0 we have x [lnx] = x. If u = u(x) is a ifferentiable function of x, then for all x such that u(x) > 0 we have x [lnu] = x [lnu(x)] = u [ u x ] = u(x) [u (x)]. Proof. We have by the Funamental Theorem of Calculus Part : x [lnx] = [ x ] x t t = x. By the Chain Rule x [lnu(x)] = u [ln u] u x = u ux. Note. We can use the above result to show that for x 0, we have x [ln x ] = x. Examples. Page 426 numbers 4 an 26.

4 7. The Logarithm Define as an Integral 4 Theorem. Properties of Logarithms. For any numbers b > 0 an x > 0 we have. lnbx = lnb + lnx 2. ln b x = lnb lnx 3. ln x = lnx 4. lnx r = r lnx. Proof. First for. Notice that [ln bx] = x bxx [bx] = bx b = x. This is the same as the erivative of lnx. Therefore by Corollary to the Mean Value Theorem, ln bx an ln x iffer by a constant, say ln bx = ln x+k for some constant k. By setting x = we nee ln b = ln +k = 0+k = k. Therefore k = lnb an we have the ientity ln bx = ln b+lnx. Now for 2. We know by : ln x + lnx = ln ( x x ) = ln = 0. Therefore ln x = lnx. Again by we have ln b ( x = ln b ) = lnb + ln = ln b lnx. x x

5 7. The Logarithm Define as an Integral 5 Finally for 4. We have by the Chain Rule: x [lnxr ] = x r x [xr ] = = r x rrxr x = r x [lnx] = [r lnx]. x As in the proof of, since ln x r an r ln x have the same erivative, we have lnx r = r lnx + k 2 for some k 2. With x = we see that k 2 = 0 an we have ln x r = r ln x. Theorem. If u is a ifferentiable function that is never zero then u = ln u + C = {ln u(x) + k k R}. u Proof. We know the result hols for u(x) > 0. We must only establish it for u(x) < 0. Notice that when u(x) < 0, u(x) > 0, an u(x) = u(x) u u = ( u) = ln( u) + C = ln u + C. u Note. We can also express the previous theorem as u(x) u (x)x = ln u(x) + C. where u(x) is nonzero.

6 7. The Logarithm Define as an Integral 6 Examples. Page 425 numbers 4 an 6. Theorem. For u = u(x) a ifferentiable function, tan u u = ln cosu + C = ln secu + C cot u u = ln sinu + C = ln cscu + C. Proof. Both follow from u-substitution: sinx tan x x = cos x x Let u = cos x then u = sinxx u u = u = u = ln u + C = ln cos x + C = ln cosx + C = ln secx + C.

7 7. The Logarithm Define as an Integral 7 Next cot x x = cos x sinx x Let u = sin x = then u = cos x x u u = ln u + C = ln sinx + C = ln = ln cscx + C. sinx The theorem follows by another application of u-substitution. Theorem. For u = u(x) a ifferentiable function, secuu = ln secu + tan u + C cscuu = ln cscu + cotu + C. Proof. Both follow from u-substitution an a trick. Here s the first

8 7. The Logarithm Define as an Integral 8 one: secxx = secx + tan x secx secx + tan x x = Let u = secx + tanx sec 2 x + secxtan x secx + tan x x = then u = (secxtan x + sec 2 x)x u = ln u + C = ln secx + tan x + C. u The result follows by another application of u-substitution. The secon result follows similarly (see page 42). Note. It is clear that the omain of the natural logarithm function is (0, ). The range of the natural logarithm function is (, ), as argue on page 420. We know that the natural logarithm function is increasing, so it is one-to-one an has an inverse. Definition. Define the natural exponential function as e x = ln x = exp x. Note. The omain of e x is (, ) (the same as the range of ln x) an the range of e x is (0, ) (the same as the omain of ln x).

9 7. The Logarithm Define as an Integral 9 Note. Since we have efine e x as the inverse of lnx, we immeiately have: e lnx = x for x (0, ) ln(e x ) = x for all x. Theorem. We have x [ex ] = e x. Proof. Let y = e x = ln x. Then Then lny = lne x = x. x [lny] = x [x] y y x = y x = y = ex. This proof iffers from that on page 422, which uses a theorem on the erivative of inverse functions (from section 3.8).

10 7. The Logarithm Define as an Integral 0 Note. By combining the previous theorem with the Chain Rule we have x [eu ] = [ ] u e u. x Example. Differentiate f(x) = e cosx secx. Theorem. We have e u u = e u + C. Proof. This is just a statement of the previous theorem in integral form. Example. Page 426 number 20. Theorem. For all numbers x, x, an x 2, the natural exponential e x obeys the following laws:. e x e x 2 = e x +x e x = e x 3. ex e x 2 = ex x 2

11 7. The Logarithm Define as an Integral 4. (e x ) x 2 = e x x 2 = (e x 2 ) x Note. The proof of Theorem is base on the efinition of y = e x in terms of x = ln y an properties of the natural logarithm function. Definition. For any numbers a > 0 an x, the exponential function with base a is given by a x = e x lna. Note. We can use the above efinition to show that Theorem hols for exponentials base a as well. Theorem. If a > 0 an u is a ifferentiable function of x, then a u is a ifferentiable function of x an x [au ] = a u lna u x. In terms of integrals, a u u = au ln a + C. Definition. For any positive number a, the logarithm of x with base a, enote by log a x, is the inverse function of a x.

12 7. The Logarithm Define as an Integral 2 Note. Since we have efine a x as the inverse of log a x, we immeiately have: a log a x = x for x (0, ) log a (a x ) = x for all x. Theorem. To convert logarithms from one base to another, we have the formula (which we may take as a efinition): or using natural logarithms: log a x = log b x log b a log a x = lnx lna. Proof. We set y = log a x an so a y = x. Taking natural logarithms, we have lna y = ln x, or y lna = lnx an y = ln x lna. Therefore, y = log a x = lnx lna. Note. We can use the previous conversion result an the properties of ln x to show that log a x satisfies all the usual properties of logarithms.

13 7. The Logarithm Define as an Integral 3 Theorem. Differentiating a logarithm base a gives: x [log a u] = ln a u ux. Proof. This follows easily: x [log a x] = x [ ] ln x lna = lnax [ln x] = ln ax. Combining this result with the Chain Rule gives the theorem. Examples. Page 426 numbers 46 an 54.

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