Find the area enclosed by the y-axis and the parametric curve $x = t^2 - 8t + 15$; $y = e^{2t}$:

Question

The title says it all. I tried this, since we need to find the area enclosed by the y-axis we could say that it is the line $x=0$, so now we can do this $t^2-8t+15 = (x-3)(x-5) = $ factorized form. Afterwards I tried the formula $A = \int_3^5 ydx$, which doesn't work. So I thought that graphing it may help. From graphic I could say that the answer may be $$\int_{-\infty}^0 e^{2t} (2t-8)dt.$$ (Thus, an improper integral), but I don't know if this is correct and it is most probably going to output a negative area. Please help.

Answer 1

The curve intersects the $y$-axis at $x=t^2-8t+15=0$, which yields $t=3$ and $t=5$. Thus, the area is

$$\int_{y_1}^{y_2} |x|dy=-\int_3^5 (t^2-8t+15)(2e^{2t})dt =\frac12 e^6(3+e^4)$$

Answer 2

It is not $(x−3)(x−5) = 0$; it is $(t−3)(t−5) = 0,$ or, $t=3,5$.

So, upper limit for $t$ is 5 and lower limit for $t$ is 3.

So, upper limit for $y =e^{2\times5} = e^{10}$ and lower limit for $y$ is $e^{2\times3} = e^6.$

Now integrate; area, $$A=\int_{e^{10}}^{e^6}xdy$$

Or, $$A=2\int_3^5(t^2-8t+15)e^(2t)dt$$

By the way, $dy = 2te^{2t}dt$.