NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Integral of a^x
Lesson

In another chapter, we looked at finding the derivative of an exponential function given by $ka^x$kax. It turns out that we need some similar thinking to find the antiderivative of an exponential function.

We saw that $\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x$ddxex=ex.

It follows that the antiderivatives are given by $\int e^x\ \mathrm{d}x=e^x+C$ex dx=ex+C

Similarly, $\frac{\mathrm{d}}{\mathrm{d}x}a^x=\ln a.e^{x\ln a}=\ln a.a^x$ddxax=lna.exlna=lna.ax

and so, it must be that $\int a^x=\frac{a^x}{\ln a}+C$ax=axlna+C.

 

Example 1

At a time $t$t seconds after the beginning of an experiment, a certain quantity $y$y is increasing at a rate given by $0.0125\times2^t.$0.0125×2t. At time $t=0$t=0, the quantity $y$y was $0.5$0.5 units. What is the value of $y$y at $t=20$t=20 seconds?

We have an expression for the rate of change of $y$y but not one for $y$y itself. We have 

$\frac{\mathrm{d}y}{\mathrm{d}t}=0.0125\times2^t$dydt=0.0125×2t

Therefore, by antidifferentiation, we have

$y$y $=$= $\int0.0125\times2^t\ \mathrm{d}t$0.0125×2t dt
  $=$= $0.0125\times\frac{2^t}{\ln2}+C$0.0125×2tln2+C

We know that $y(0)=0.5$y(0)=0.5. So, $0.0125\times\frac{2^0}{\ln2}+C=0.5$0.0125×20ln2+C=0.5 and therefore, $C=0.5-0.0125\times\frac{2^0}{\ln2}\approx0.482$C=0.50.0125×20ln20.482.

So, the rule for $y$y itself is $y(t)=0.0125\times\frac{2^t}{\ln2}+0.482$y(t)=0.0125×2tln2+0.482. From this, we obtain

$y(20)=0.0125\times\frac{2^{20}}{\ln2}+0.482\approx18910$y(20)=0.0125×220ln2+0.48218910 units

 

 

Example 2

In Statistics, we say that a continuous non-negative function defined over an interval can serve as a probability density function if its integral over the interval is one.

For example, the function $f(x)=\frac{1}{3}$f(x)=13 defined on the interval $[1,4]$[1,4] is a probability density function because $\int_1^4\left(\frac{1}{3}\right)\ \mathrm{d}x$41(13) dx evaluates to $1$1.

Consider the exponential function $g(t)=1.0986\times3^{-t}$g(t)=1.0986×3t defined on the interval $[0,\infty)$[0,). Show that it could serve as a probability density function.

We have to show that $\int_0^{\infty}1.0986\times3^{-t}\ \mathrm{d}t=1$01.0986×3t dt=1.

We should pause for a moment to think about the effect of the negative exponent. By the function-of-a-function rule, we have $\frac{\mathrm{d}}{\mathrm{d}t}e^{-t}=-e^{-t}$ddtet=et. Also, if we want  $a^{-t}=e^y$at=ey, then $y=-t\ln a$y=tlna by the same reasoning as was used above.

It follows, much as before, that

$\frac{\mathrm{d}}{\mathrm{d}t}a^{-t}$ddtat $=$= $\frac{\mathrm{d}}{\mathrm{d}t}e^{-t\ln a}$ddtetlna
  $=$= $-\ln a.e^{-t\ln a}$lna.etlna
  $=$= $-\ln a.a^{-t}$lna.at

And then, it is not difficult to confirm by differentiation that $\int a^{-t}\ \mathrm{d}t=-\frac{a^{-t}}{\ln a}+C$at dt=atlna+C.

Returning to the problem, we need to evaluate $\int_0^{\infty}1.0986\times3^{-t}\ \mathrm{d}t$01.0986×3t dt

This is, $1.0986\left[-\frac{3^{-t}}{\ln3}\right]_0^{\infty}$1.0986[3tln3]0 
$=1.0986\left[0--\frac{1}{\ln3}\right]$=1.0986[01ln3]

$\approx1$1

as required.

 

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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