Integrating Parametric Equations: Summary

Here’s a reminder of the key points you should know about integrating parametric equations.

If we’re given a curve that’s defined parametrically…[x=t ,    y=t ,  0≤t≤π ]

And asked to integrate y with respect to x…

It might be possible to just convert the parametric equations into cartesian form first.[y=1x ]

And then integrate as normal.[∫y dx=ln x +c](pause for 1 sec)

If we want to avoid converting parametric equations to cartesian form before integrating…[x=3t+4,  y=4cos t , 0≤t≤2]

We can just take our equation for y…[highlight y=4cos t ]

And take the integral of both expressions with respect to x…[y dx=4cos t dx]

Then, take our equation for x…[x=3t+4]

Differentiate… [dxdt=3]

And make this the subject, so we can convert “dx”.[dx=3dt, plug it into the integral so we get:y dx=4cos t (3) dt ]

Finally, expand and integrate as normal.

[y dx=12cos t  dt]

[y dx=12sin t +c](pause for 1 sec)

And if we now want to find an area under the curve…[between 4 and ]

We could rewrite our integral in terms of x. [t=4x-π12, plug into the result to get y dx=12sin 4x-π12 ]

…And evaluate it between the boundaries of our area. [4y dx=[12sin 4x-π12 ]4=62](pause for 1 sec)

However, the smartest thing to do will almost definitely be to convert the boundaries [highlight 4 and ] into t values instead. [take the formula t=4x-π12 and replace the boundaries with t=4(4)-π12 and t=4()-π12, and morph into 0 and 4]