I don't understand how to obtain the limits for the $t$-values considering that they gave us the $x$-values in the first part of the equation. I've considered substituting the $x$-values into the parametric $x$ equation, and trying to obtain $t$ in that way, but obviously I would obtain 3 $t$-values, after which I have no idea how to compute, as all my integration would be in terms of $t$.

Here's the question:

A curve is defined parametrically by $x=t^3-3t+2$ and $y=3(t^2-1)$ where $t \in \mathbb{R}$.


i) the mean value of $y$ with respect to $x$ for $1$ $\leq$ $x$ $\leq$ $2$

ii) the length of the curve from the origin $O$, where $t$ = $1$, to the point $A$ where $t$ = $2$

iii) the area of the surface generated when the arc $OA$ is rotated through one revolution about the $x$-axis, leaving your answer in terms of $\pi$

This is my progress

$$dx/dt = 3t^2-3$$

$$y_m = \frac{1}{x_2-x_1}9\int_{t_1}^{t_2} (t^4-2t^2+1) \, \mathrm{d}t$$

My main problem is finding $t_1$ and $t_2$. Thanks for your help.

  • 2
    $\begingroup$ You can find $t_1$ and $t_2$ if you solve equations $x(t_1)=1$ and $x(t_2)=2$ $\endgroup$ – cool Aug 28 '14 at 14:56
  • $\begingroup$ Shouldn't be $t_2 - t_1$ in the denominator in the expression of $y_m$? $\endgroup$ – Dmoreno Aug 28 '14 at 15:06

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