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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.

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enter image description here

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

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

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  • $\begingroup$ Can you please explain where the - next to the integration symbol came from? $\endgroup$ – Arman Babayan Mar 30 '20 at 19:37
  • $\begingroup$ @ArmanBabayan - note that $x=t^2 -8t+15$ is negative over $t\in [3,5]$. The ‘-‘ ensures that the integrand for the area is positive $\endgroup$ – Quanto Mar 30 '20 at 19:42
  • $\begingroup$ Thank you so much, you are the only person, whose solution was clear to me. If you please could, could you look at another question I posted here about the same topic? math.stackexchange.com/questions/3601479/… $\endgroup$ – Arman Babayan Mar 30 '20 at 19:45
  • $\begingroup$ @ArmanBabayan - no problem. I’ll take a look $\endgroup$ – Quanto Mar 30 '20 at 20:10
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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, $dy = 2te^{2t}dt$.

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  • $\begingroup$ I think you mixed the limits of integration position in the first integral, also dy = 2e^(2t)dt. But I don't understand one thing: Why do these limits of integration yield to the same thing? How is e^6 = 3 and e^10 = 5???? $\endgroup$ – Arman Babayan Mar 30 '20 at 19:39

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