The curve C has parametric equations $x = n + 2$, $y = rac{1}{(t + 1)}$, $t > -1$ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 8

To evaluate the integral,

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

we can perform partial fraction decomposition:

$\frac{1}{(t + 1)(t + 2)} = \frac{A}{(t + 1)} + \frac{B}{(t + 2)}$.

By equating, we find:

• Multiply through by the denominator:

$1 = A(t + 2) + B(t + 1)$.

Solving for A and B gives:

• A = 1 and B = -1.

Consequently, we rewrite the integral:

$\int_{0}^{2} \left(\frac{1}{(t + 1)} - \frac{1}{(t + 2)}\right) dt$.

Now, integrate:

$\left[\ln(t + 1) - \ln(t + 2)\right]_{0}^{2} = (\ln(3) - \ln(4)) - (\ln(1) - \ln(2))$

Therefore, simplifying the expression gives:

$A = \ln(\frac{3}{4}) - \ln(1) + \ln(2) = \ln(\frac{3}{2})$.

Hence, the exact value of the area is:

$A = \ln(\frac{3}{2})$.

To find the cartesian equation from the parametric equations:

1. Given:
   $x = \ln(t + 2)$
   $y = \frac{1}{(t + 1)}$

2. Rearranging for t from the equation of x:
   $t + 2 = e^{x} \\ t = e^{x} - 2$

3. Substitute t into the equation for y: $y = \frac{1}{(e^{x} - 1)}$

Thus, the cartesian equation of the curve in the form $y = f(x)$ is:

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

The domain for x must ensure that the denominator is not zero:

From the equation $y = \frac{1}{(e^{x} - 1)}$,

• We need $e^{x} - 1 > 0$ ⇔ $e^{x} > 1$.
• This implies $x > 0$.

Thus, the domain of values for x is:

$x > 0$.