Question

in 2010 a particular factory produced computer hard drives at a rate of ( r(t)=4.4e^{0.0175t} ) thousand drives per month and ( g(t)=5.4e^{0.0163t} ) thousand drives per month where ( t ) is the time in months after the start of that year. in 2011 the production rate was ( t ) months after the start of that year. compute the area between the graphs of these functions for ( 0leq tleq12 ). (round your answer to three decimal places.)

Explanation:

Step1: Recall the formula for the area between two curves

The area $A$ between two curves $y = f(t)$ and $y = g(t)$ from $t=a$ to $t = b$ is given by $A=\int_{a}^{b}|f(t)-g(t)|dt$. Here, $f(t)=4.4e^{0.0175t}$, $g(t)=5.4e^{0.0163t}$, $a = 0$, and $b = 12$.

Step2: Set up the integral

$A=\int_{0}^{12}(5.4e^{0.0163t}-4.4e^{0.0175t})dt$ (since for $t\in[0,12]$, $5.4e^{0.0163t}\geq4.4e^{0.0175t}$).

Step3: Integrate term - by - term

The integral of $e^{kt}$ with respect to $t$ is $\frac{1}{k}e^{kt}+C$.
$\int_{0}^{12}5.4e^{0.0163t}dt=5.4\times\frac{1}{0.0163}e^{0.0163t}\big|_{0}^{12}=\frac{5.4}{0.0163}(e^{0.0163\times12}-e^{0})$
$\int_{0}^{12}4.4e^{0.0175t}dt=4.4\times\frac{1}{0.0175}e^{0.0175t}\big|_{0}^{12}=\frac{4.4}{0.0175}(e^{0.0175\times12}-e^{0})$

Step4: Calculate the values

$\frac{5.4}{0.0163}(e^{0.1956}-1)-\frac{4.4}{0.0175}(e^{0.21}-1)$
$e^{0.1956}\approx1.2167$, $e^{0.21}\approx1.2337$
$\frac{5.4}{0.0163}(1.2167 - 1)-\frac{4.4}{0.0175}(1.2337 - 1)$
$=\frac{5.4}{0.0163}\times0.2167-\frac{4.4}{0.0175}\times0.2337$
$=\frac{5.4\times0.2167}{0.0163}-\frac{4.4\times0.2337}{0.0175}$
$=\frac{1.17018}{0.0163}-\frac{1.02828}{0.0175}$
$\approx71.790 - 58.759$
$=13.031$

Answer:

$13.031$