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This is one of the trickiest methods of calculus on the course, but it's important to know, and is very doable if you set up the problem right and remember the steps.
Integration by part...
Every point on the curve C satisfies the equation. In order to show P lies on C, we need to test if either x- or y-coordinates satisfy the equation. It is easier to subsitute x=2 into the equation.
Starting with the initial integral of int(exp(x).cos(x))dx we can see that this is going to have to be integrated by parts. This states that the integral of (u . dv/dx)dx is equal to u.v - int(v . du/d...
To find the gradient of any curve, we take the derivative. So in this case, we need to take dy/dx. We do this by multiplying the term by the power on x, and then lowering the power by one. For example, fo...
Let u = x and dv/dx = sin(x),
By using the general expression of:
integral(u multiply dv/dx)dx = [u multiply v] - integral(v multiply du/dx)dx, and by realising that:
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