Answer:
The additional amount of area covered between 4 to 9 days is 23.71 cm2
Step-by-step explanation:
As the relation is given as a combination of two functions so integration by parts is carried out thus
[tex]\int\limits^9_4 {\sqrt{t}\, ln t} \, dt[/tex]
In order to solve this integral, integration by parts is to be carried out which is given as
[tex]\int u v dx=u \int v dx -\int u'(\int vdx) dx[/tex]
Where
u(x) is a function of x
v(x) is a function of x
u' is the derivative of u wrt to x
Also u and v are defined on using the following sequence ILATE RULE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)
As here Logarithmic function is present which is taken as u and the algebraic function is taken as v so
[tex]u= ln t\\v=\sqrt{t}\\u'=\frac{1}{t}[/tex]
[tex]\int v dt =\int t^{1/2} dt =\frac{2}{3}t^{3/2}[/tex]
Substituting the values in equation gives
[tex]\int u v dt=u \int v dt -\int u'(\int vdt) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{1}{t} )(\frac{2}{3}t^{3/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{2}{3}t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}\int(t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}(\frac{2}{3}t^{3/2}) +C\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C[/tex]
Now solving the definite integral
[tex]\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{9} -[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{4}\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=18 ln (9)-\frac{16}{3} ln 4 -\frac{76}{9}\\A=\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=23.71 cm^2\\[/tex]
So the additional amount of area covered between 4 to 9 days is 23.71 cm2
