Find the area enclosed by the line $x=5$ and the parametric curve $x=6t-t^2, y=e^{3t}$

Question

Well, I have written the question above. This is what I tried, even though all of this is quite messed up, and I’m sorry for that. Could anybody guide me through the complete solution? Can you just solve it down here? I’ll understand if I see the solution. ]2]2

Answer 1

The curve intersects with the vertical boundary $x=5$ at $x=6t-t^2=5$, which yields $t=1$ and $t=5$. Note that the area is to the right of the boundary $x=5$, which is shaded in the graph.

Thus, the area integral is given by

$$\int_{y_1}^{y_2} (x-5)dy=\int_1^5 (6t-t^2-5)(3e^{3t})dt =\frac29 e^3(7+5e^{12})$$

Answer 2

We have to find the area bounded by $x=5$ and $$y=\exp(9-3\sqrt{9-x})$$ How we reached here:
$$x=6t-t^2$$ $$ t^2-6t+x=0$$ This is a quadratic in $t$. Applying the formula and solving for $t$ , we'll get: $$ t= \dfrac{6\pm\sqrt{36-4x}}{2}$$ $$\Rightarrow t=3\pm\sqrt{9-x}$$ Plugging this $t$ in the parametric equation of $y$, we get: $$y=\exp(9 \pm 3\sqrt{9-x})$$ We consider only $$y=\exp(9-3\sqrt{9-x})$$ The other equation we get by solving for the explicit curve is ignored because of this graph : Assuming we want the area bounded from $-\infty$ to the point where the curve cuts the line $x=5$ , we have to perform the following integral : $$ \int\limits_{-\infty}^{5}\exp({9-3\sqrt{9-x}}) \ \mathrm{ d}x$$

Solving this (pretty calculative) integral ends us up with : $$\dfrac{2\left(3\sqrt{9-x}+1\right)\mathrm{e}^{9-3\sqrt{9-x}}}{9}$$ And upon placing limits, The area bound turns out to be: $$\dfrac{14\mathrm{e}^3}{9}$$

Answer 3

$A = \int y\ dx\\ y = e^{3t}\\ x = 6t -t^2\\ dx = 6 - 2t \ dt\\ \int e^{3t} (6-2t)\ dt$

limits:

$5 = 6t - t^2\\ (t-1)(t-5) = 0$

$\int_1^5 -e^{3t}(6-2t)\ dt$

$-2e^{3t} + \frac {2t e^{3t}}{3} - \frac {2e^{3t}}{9}$

$\frac {(-20+6t)e^{3t}}{9}|_1^5\\ \frac {10e^{15} + 14e^{3}}{9}$