Respuesta :
a.) 0.5367 feet
b.) 0.5223 feet'
Given the rate at which an eucalyptus tree will grow modelled by the equation 0.5+6/(t+4)³ feet per year, where t is the time (in years).
The amount of growth can be gotten by integrating the given rate equation as shown:
[tex]\int\limits {0.5} \, + \frac{6}{(t+4)^{3} } dt[/tex]
[tex]=\int\limits {0.5} \,dt + \int\limits\frac{6}{(t+4)^{3} } dx[/tex]
[tex]=0.5t+\int\limits6u^{-3} du \, \,where \,u=t +4 \, and \, du=dt[/tex]
[tex]=0.5t+6*\frac{u^{-3} }{-2} + C[/tex]
[tex]=0.5t-3u^{-2}+C[/tex]
[tex]=0.5t-3(t+4)^{-2} +C[/tex]
a) The number of feet that the tree will grow in the second year can be gotten by taking the limit of the integral from t =1 to t = 2
[tex]\int\limits^2_1 {0.5} \,dt + \int\limits\frac{6}{(t+4)^{3} } dx \, =[0.5t-3(t+4)^{-2} ][/tex]
[tex]=[0.5(2)-3(2+4)^{-2} ]-[0.5-3(5)^{-2} ][/tex]
[tex]=\frac{11}{12}-\frac{1}{2}-\frac{3}{25}[/tex]
[tex]=0.9167-0.5+0.12[/tex]
[tex]=0.5367[/tex]
b) The number of feet that the tree will grow in the third year can be gotten by taking the limit of the integral from t =2 to t = 3
[tex]\int\limits^3_2 {0.5} \,dt + \int\limits\frac{6}{(t+4)^{3} } dx \,[/tex]
[tex]=[0.5t-3(t+4)^{-2} ][/tex]
[tex]=[1.5-\frac{3}{49} ]-[1-\frac{1}{12} ][/tex]
=1.439-0.9167
=0.5223 feet
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