Show that the area of R is given by the integral ∫(2, 0) (1 / ((t + 1)(t + 2))) dt.

Steps to Solve

1. Identify the area under the curve

The area ( A ) under the curve ( C ) from ( x = \ln 2 ) to ( x = \ln 4 ) can be expressed as:

$$ A = \int_{a}^{b} y , dx $$

where ( a = \ln 2 ) and ( b = \ln 4 ).

2. Change of variables

Since the parametric equations are given by ( x = \ln(t + 2) ) and ( y = \frac{1}{t + 1} ), we first need to express ( y ) in terms of ( t ). We can relate ( t ) to ( x ) using the equation for ( x ):

$$ e^x = t + 2 \implies t = e^x - 2 $$

Substituting this into the equation for ( y ):

$$ y = \frac{1}{(e^x - 2) + 1} = \frac{1}{e^x - 1} $$

3. Finding the derivative

Now we need to find ( \frac{dx}{dt} ) to convert the integral. We have ( x = \ln(t + 2) ):

$$ \frac{dx}{dt} = \frac{1}{t + 2} $$

Then, invert this to get ( dx ):

$$ dx = (t + 2) , dt $$

4. Set the limits for ( t )

Now we need to establish the limits of integration in terms of ( t ). When ( x = \ln 2 ):

$$ 2 = t + 2 \implies t = 0 $$

When ( x = \ln 4 ):

$$ 4 = t + 2 \implies t = 2 $$

Thus, the new limits of integration are from ( t = 0 ) to ( t = 2 ).

5. Substituting into the integral

The area ( A ) can now be expressed as:

$$ A = \int_{0}^{2} y \cdot \frac{dx}{dt} , dt $$

Substituting ( y = \frac{1}{t + 1} ) and ( \frac{dx}{dt} = (t + 2) ):

$$ A = \int_{0}^{2} \frac{1}{t + 1} \cdot (t + 2) , dt $$

6. Final integral expression

This gives us:

$$ A = \int_{0}^{2} \frac{1}{(t + 1)(t + 2)} , dt $$

Thus, we have shown that the area of region ( R ) is given by the integral:

$$ A = \int_{0}^{2} \frac{1}{(t + 1)(t + 2)} , dt $$