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.
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=0.5−0.0125×20ln2≈0.482.
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+0.482≈18910 units
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)=1.0986×3−t defined on the interval $[0,\infty)$[0,∞). Show that it could serve as a probability density function.
We have to show that $\int_0^{\infty}1.0986\times3^{-t}\ \mathrm{d}t=1$∫∞01.0986×3−t 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 $\frac{\mathrm{d}}{\mathrm{d}t}e^{-t}=-e^{-t}$ddte−t=−e−t. Also, if we want $a^{-t}=e^y$a−t=ey, then $y=-t\ln a$y=−tlna by the same reasoning as was used above.
It follows, much as before, that
$\frac{\mathrm{d}}{\mathrm{d}t}a^{-t}$ddta−t | $=$= | $\frac{\mathrm{d}}{\mathrm{d}t}e^{-t\ln a}$ddte−tlna |
$=$= | $-\ln a.e^{-t\ln a}$−lna.e−tlna | |
$=$= | $-\ln a.a^{-t}$−lna.a−t |
And then, it is not difficult to confirm by differentiation that $\int a^{-t}\ \mathrm{d}t=-\frac{a^{-t}}{\ln a}+C$∫a−t dt=−a−tlna+C.
Returning to the problem, we need to evaluate $\int_0^{\infty}1.0986\times3^{-t}\ \mathrm{d}t$∫∞01.0986×3−t dt.
This is, $1.0986\left[-\frac{3^{-t}}{\ln3}\right]_0^{\infty}$1.0986[−3−tln3]∞0
$=1.0986\left[0--\frac{1}{\ln3}\right]$=1.0986[0−−1ln3]
$\approx1$≈1
as required.
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
Apply integration methods in solving problems