# Thread: Integration question involving parametric equations

1. ## Integration question involving parametric equations

The curve C has parametric equations $x=\ln(t+2), y=\frac{1}{(t+1)}, t >-1$

The finite region R between the curve C and the x-axis is bounded by the lines with equations x = ln 2 and x = ln 4.

a)Show that the area of R is given by the integral $\int_{0}^{2}\frac{1}{(t+1)(t+2)}\,dt$ (4 marks)

b)Hence find an exact value for this area (6 marks)

c)Find a cartesian equation of the curve C, in the form y=f(x) (4 marks)

d)State the domain of values for x for this curve (1 mark)

I came across this question on a past exam paper i was using to revise from and although i attempted it I am still quite unsure of how to tackle this question or a similar question if it comes up.

I would be grateful if somebody could show me how to go about answering this question and give me some guidelines of how to think about a question like this in the future. Thanks to all!

2. The thing to remember for these types of questions is that the region you require is

$\int^a_b y\, dx$

$= \int^a_b y \frac{dx}{dt} dt$

by the chain rule.

so for you we have

$\int^a_b \frac{1}{(t+1)} \frac{dx}{dt} dt = \int^a_b \frac{1}{(t+1)} \frac{1}{(t+2)} dt$

Now you need to find your values of a and b, well they are when x = ln(4) and x = ln(2), look at your expression for x and you see that these occur when t = 2 and 0 respectively.

So putting it all together we have

$R = \int^2_0 \frac{1}{(t+1)(t+2)} \, dt$

as required, which you can solve using partial fractions

3. Originally Posted by Big_Joe
The curve C has parametric equations $x=\ln(t+2), y=\frac{1}{(t+1)}, t >-1$

The finite region R between the curve C and the x-axis is bounded by the lines with equations x = ln 2 and x = ln 4.

a)Show that the area of R is given by the integral $\int_{0}^{2}\frac{1}{(t+1)(t+2)}\,dt$ (4 marks)

b)Hence find an exact value for this area (6 marks)
he also addressed this. use partial fractions...or as an alternative, algebra

Note that $\frac 1{(t + 1)(t + 2)} = \frac {t + 2 - (t + 1)}{(t + 1)(t + 2)} = \frac {t + 2}{(t + 1)(t + 2)} - \frac {t + 1}{(t + 1)(t + 2)} = \frac 1{t + 1} - \frac 1{t + 2}$

this is where partial fractions would have gotten you. now, integrate term by term

c)Find a cartesian equation of the curve C, in the form y=f(x) (4 marks)
look at $x = \ln (t + 2)$

$\Rightarrow t + 2 = e^x$

$\Rightarrow t + 1 = e^x - 1$

thus, we can replace the $t + 1$ in the $y(t)$ function to get the $y(x)$ function we require.

d)State the domain of values for x for this curve (1 mark)
i suppose you can handle this, bearing in mind how our curve is defined and the answer to part (c)