NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Integration to give log functions
Lesson

Introduction

Having ascertained that, for the function $y=\ln x$y=lnx, the derivative $\frac{dy}{dx}=\frac{1}{x}$dydx=1x, then it is also true that;

$\int\frac{1}{x}dx=\ln x+C$1xdx=lnx+C

In fact it is more correct to say that $\int\frac{dx}{x}=\ln\left|x\right|+C$dxx=ln|x|+C, reminding us that any argument of a log function must always be positive.

However, for the purposes of this discussion, we will omit the absolute value signs as is the regular practice of most school text books on this topic. 

 

Two observations

The result is interesting because it fills in the gap that became apparent in one of the first  general rules of integration - namely that $\int x^ndx=\frac{x^{n+1}}{n+1},n\ne-1$xndx=xn+1n+1,n1

It is mysterious in that there is a link between the area under the rectangular hyperbola and the natural logarithm.

Thus we find that $\int_1^a\frac{dx}{x}=\left[\ln x\right]_1^a=\ln a$a1dxx=[lnx]a1=lna as shown here:

 

As an interesting consequence of this result, the open area between the $x$x and $y$y axes and the upper right arc of the curve given by $y=\frac{1}{x}$y=1x can be broken up into unit-size portions bounded by integer powers of $e$e as shown in this schematic diagram: 

 

For example the area under the curve between $x=e$x=e and $x=e^2$x=e2 is evaluated as:

$\int_e^{e^2}\frac{1}{x}dx=\left[\ln x\right]_e^{e^2}=2\ln e-\ln e=1$e2e1xdx=[lnx]e2e=2lnelne=1

Because of the symmetry of $y=\frac{1}{x}$y=1x, the same result will apply with respect to either axis.

 

Estimate of log 2 

method 1

As mentioned above it seems mysterious that for all values of $n\ne-1$n1 we have $\int x^ndx=\frac{x^{n+1}}{n+1}+C$xndx=xn+1n+1+C, but somehow, when $x=-1$x=1, there is an abrupt change in solution.

If we think about the curve of the function $y=x^n$y=xn for values of $n$n close to (but not equal to) $-1$1, we might expect that evaluated areas under such curves would be approximately equal to the area under the curve when $n$n is precisely $-1$1.

For example, we would expect that $\int_1^2x^{-0.99}dx\approx\int_1^2x^{-1}dx=\ln2$21x0.99dx21x1dx=ln2.

Evaluating the first of these integrals we have:

$\int_1^2x^{-0.99}dx$21x0.99dx $=$= $\left[\frac{x^{-0.99+1}}{-0.99+1}\right]_1^2$[x0.99+10.99+1]21
  $=$= $\left(\frac{x^{0.01}}{0.01}\right)-\left(\frac{1}{0.01}\right)$(x0.010.01)(10.01)
  $=$= $100\left(\sqrt[100]{2}-1\right)$100(10021)
     

The value $100\left(\sqrt[100]{2}-1\right)\approx0.695555$100(10021)0.695555 which is quite close to $\ln2\approx0.693147$ln20.693147.

Had we chosen $n$n even closer to $-1$1, we would have arrived at a solution closer to $\ln2$ln2

In fact, in keeping with the pattern above, an integral like $\int_1^2x^{-0.9999}dx$21x0.9999dx would evaluate to $10000\left(\sqrt[10000]{2}-1\right)\approx0.693171$10000(1000021)0.693171.

This leads naturally to the understanding that $\ln2=\lim_{k\rightarrow\infty}k\left(\sqrt[k]{2}-1\right)$ln2=limkk(k21) and in an even more general statement $\ln a=\lim_{k\rightarrow\infty}k\left(\sqrt[k]{a}-1\right)$lna=limkk(ka1).

 

Applet

The applet below evaluates the integral given by $\int_1^ax^ndx$a1xndx for $-1.011.01<n<0.99 allowing you to very slowly increment $n$n (in increments of $0.0001$0.0001) for various values of a. When $n=-1$n=1, the value that is shown switches to $\ln a$lna

 
historical note

As an historical note, if we momentarily set $k(\sqrt[k]{a}-1)=\ln a$k(ka1)=lna, we can rearrange as follows:

$k(\sqrt[k]{a}-1)$k(ka1) $=$= $\ln a$lna
$\sqrt[k]{a}-1$ka1 $=$= $\frac{\ln a}{k}$lnak
$\sqrt[k]{a}$ka $=$= $1+\frac{\ln a}{k}$1+lnak
$a$a $=$= $\left(1+\frac{\ln a}{k}\right)^k$(1+lnak)k
     

Of course this is only true in the limit as $k$k approaches infinity.

By setting $a=e$a=e, and using limits, what we have actually deduced is Leonard Euler's famous discovery that:

$\lim_{k\rightarrow\infty}\left(1+\frac{1}{k}\right)^k=e$limk(1+1k)k=e.

 

Estimate of log2

METHOD 2

Using an alternative approach the following sketch depicts ten circumscribed rectangles drawn on the curve of $y=\frac{1}{x}$y=1x between $x=1$x=1 and $x=2$x=2 in order to gain an upper bound estimate of $\ln2$ln2 (we could also use inscribed rectangles to get a lower bound estimate).

  

With 10 rectangles, we readily see that the area is given by $A_{10}$A10 where:

$A_{10}$A10 $=$= $\frac{1}{10}\left(\frac{1}{1}+\frac{1}{1+\frac{1}{10}}+\frac{1}{1+\frac{2}{10}}+...+\frac{1}{1+\frac{9}{10}}\right)$110(11+11+110+11+210+...+11+910)
  $=$= $\frac{1}{10}\left(\frac{10}{10}+\frac{10}{11}+\frac{10}{12}+...+\frac{10}{19}\right)$110(1010+1011+1012+...+1019)
  $=$= $\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{19}\right)$(110+111+112+...+119)
  $=$= $0.7187714...$0.7187714...
     

With more rectangles, we obtain a closer estimate of $\ln2$ln2. Using a spreadsheet and the formula given by $A_n=\left(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}\right)$An=(1n+1n+1+1n+2+...+12n1) a value of $0.69331388$0.69331388 was obtained for $n=1500$n=1500

Using inscribed rectangles with the formula $A_n=\left(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}\right)$An=(1n+1+1n+2+1n+3+...+12n) a value of $0.69298054$0.69298054 was obtained for the same number of rectangles.

The average of these two values, given as $\frac{0.69331388+0.69298054}{2}=0.69314721$0.69331388+0.692980542=0.69314721 compares extremely well with $\ln2=0.69314718....$ln2=0.69314718.....

 

Worked Examples

Question 1

The derivative of $f\left(x\right)$f(x) is $\frac{1}{x}$1x.

Which of the following could be the function? Select all the correct options.

  1. $f\left(x\right)=k\ln x$f(x)=klnx

    A

    $f\left(x\right)=\ln\left(\left|kx\right|\right)$f(x)=ln(|kx|)

    B

    $f\left(x\right)=\ln kx$f(x)=lnkx for $k<0$k<0, $x>0$x>0

    C

    $f\left(x\right)=\ln x$f(x)=lnx

    D

    $f\left(x\right)=\ln kx$f(x)=lnkx for $k>0$k>0, $x<0$x<0

    E

    $f\left(x\right)=k\ln x$f(x)=klnx

    A

    $f\left(x\right)=\ln\left(\left|kx\right|\right)$f(x)=ln(|kx|)

    B

    $f\left(x\right)=\ln kx$f(x)=lnkx for $k<0$k<0, $x>0$x>0

    C

    $f\left(x\right)=\ln x$f(x)=lnx

    D

    $f\left(x\right)=\ln kx$f(x)=lnkx for $k>0$k>0, $x<0$x<0

    E

Question 2

Determine $\int\frac{3}{4x}dx$34xdx.

Use $C$C as the constant of integration.

Question 3

Find the exact value of $\int_4^6\frac{1}{x-2}dx$641x2dx.

Express your answer in simplest form.

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91579

Apply integration methods in solving problems

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