(c) Find the area contained between the curve, the y-axis, the line t = 1 and the asymptote to the curve which is parallel to the t-axis.

Question

Part (a) and (b) are fine, and I believe (c) is an integral, but i'm not quite sure how to go about solving said integral with the given parameters. Mainly the vertical limits, as i'm sure the horizontal limits are just 0 and 1

Answer 1

Here, it's given that

$$f(t) = \frac{t^2 + 3t + 3}{(t+1)^2} = 1 + \frac{t+2}{(t+1)^2}$$

Now, as $t \to \infty$, $f(t) \to 1$, as the square term in the denominator would decay faster than the numerator term

Hence, you need to find the area under the curve, but above $f(t)=1$, and with $0 \leq t\leq 1$

Now, $f(1) = 7/4 > 1$ and $f$ is monotonically decreasing (check with derivative)

Hence

$$A= \int_0^1f(t)dt - 1$$

Answer 2

As it is clear from the graph that you have to find $\displaystyle\int_{0}^1 (f(t)-1)dx$ . You can simplify $f(t)$ as $$f(t)=1+\frac{1}{t+1}+\frac{1}{(t+1)^2}$$