New Zealand
Level 8 - NCEA Level 3

# 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:

 $\int_1^2x^{-0.99}dx$∫21​x−0.99dx $=$= $\left[\frac{x^{-0.99+1}}{-0.99+1}\right]_1^2$[x−0.99+1−0.99+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(100√2−1)

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).

### 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