so the area is basically a trapezoid of sorts; another way of putting it would be depicting the problem as a function f(x) = ax + b, and asking what's the integral from 2 to 1 of f(x) dx. or, replacing y with f(x) ∫y dx (2 to 1)
This is the first step in your solution.
The problem here is that we have parametric coordinates. Y is defined in terms of t, so ∫ydx is unsolvable as it is (we don't know what y with respect to x is). So in order for the integral ydx to be solvable, we have to either substitute dx for t, or substitute y in terms of t. In the above situation, they chose to substitute dx for t. Here's the math they left out:
x=2t5
dx/dt=2
dx=2dt
So...
∫ydx (x = 2 to 1)
turn into (just plug in 2dt for dx)
∫y*dx/dt*dt = ∫y*2dt (as x = 2 to 1).
We now have an integral with respect to t, but with bounds with respect to x.
To get to the second step, we need to figure out the bounds on t as x goes 2 to 1.
We do this like this:
x=2t5
when x = 2
3=2t, t = 3/2
when x = 1
4=2t, t = 2
So the integral needs to be rewritten as follows:
∫y*2*dt (as t goes from 3/2 to 2)
which equals this.
∫2ydt (as t goes 3/2 to 2)
So the final solution would be:
∫2ydt (as t goes 3/2 to 2) =
∫2*(t1) dt (blah)=
... simple calculus from there, i think.
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