Answer:
a) [tex]f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000[/tex]
b) f(t=2015) = 264,034,317.7
Step-by-step explanation:
The rate of change in the number of hospital outpatient visits, in millions, is given by:
[tex]f'(t)=0.001155t(t-1980)^{0.5}[/tex]
a) To find the function f(t) you integrate f(t):
[tex]\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt[/tex]
To solve the integral you use:
[tex]\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}[/tex]
Next, you replace in the integral:
[tex]\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C[/tex]
Then, the function f(t) is:
[tex]f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'[/tex]
The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.
Hence C' = 264,034,000
The function is:
[tex]f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000[/tex]
b) For t = 2015 you have:
[tex]f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7[/tex]
