Question

The number of research articles in a prominent journal written by researchers during 1983–2003 can be approximated by U(t) = 4.1e0.6t 0.2 + e0.6t thousand articles per year (0 ≤ t ≤ 20) where t is time in years (t = 0 represents 1983). (a) Find an (approximate) expression for the total number of articles (in thousands) written since 1983 (t = 0). HINT [See Example 5.] (Round your coefficients to two decimal places.) N(t) = (b) Roughly how many articles were written from 1983 to 2003? (Round your answer to the nearest whole number.) thousand articles

Answer 1

Answer:

Step-by-step explanation:

From the information given:

We are to find; an expression for the total no. of articles written since 1983

∴

The total no. of articles written since 1983  $=\int \limts ^t_0 U(t) dt$

$= \int ^t_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^t_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{0.6 x}}{1.2} \bigg|\bigg]$

$=\dfrac{41}{6} \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-\dfrac{41}{6} \mathtt{ In} \bigg | 1.2\bigg |$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.2458$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

Therefore, the total number of articles written since 1983 is  $=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

b. To find how many articles were being written  from 1983 to 2003

i.e. t = 2003 - 1983 = 20

∴

Total articles written from 1983 to  2003 is $=\int \limts ^{20}_0 U(t) dt$

$= \int ^{20}_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^{20}_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 *20} \bigg| - \mathtt{In} \bigg|0.2+e^{0.6*0}\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+e^0\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|1.2\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{12}}{1.2} \bigg|\bigg]$

= 80.75 thousand articles