Question

APPLY IT Rate of Growth The area covered by a patch of moss is growing at a rate of

Answer 1

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

$\int\limits^9_4 {\sqrt{t}\, ln t} \, dt$

In order to solve this integral, integration by parts is to be carried out which is given as

$\int u v dx=u \int v dx -\int u'(\int vdx) dx$

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

$u= ln t\\v=\sqrt{t}\\u'=\frac{1}{t}$

$\int v dt =\int t^{1/2} dt =\frac{2}{3}t^{3/2}$

Substituting the values in equation gives

$\int u v dt=u \int v dt -\int u'(\int vdt) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{1}{t} )(\frac{2}{3}t^{3/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{2}{3}t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}\int(t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}(\frac{2}{3}t^{3/2}) +C\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C$

Now solving the definite integral

$\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{9} -[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{4}\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=18 ln (9)-\frac{16}{3} ln 4 -\frac{76}{9}\\A=\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=23.71 cm^2$

So the additional amount of area covered between 4 to 9 days is 23.71 cm2