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The first step is to factorise the equation into two brackets. In this case we get (2x+1)(x-6)Now, for this to be greater than zero we need both brackets to be greater than zero, or both brackets to be le...

First, find the centre of the circle. Since we know the equation of the circle, we know that the centre (-g,-f) comes from the part of the equation -12x-6y, where -12=2g and -6=2f. Therefore, to get the c...

y = 3x⁴-8x³-3dy/dx = (3 x 4)x³-(8 x 3)x² dy/dx = 12x³ - 24x²

The first step is deciding on the method of integration. For this integral it makes the most sense to use substitution.Let u = x² + 1Differentiate w.r.t x => du/dx = 2xRearrange for dx=> ...

Integrate by parts:u = ln(x) u' = 1/x v' = 1 v = xBy parts formula: uv - ∫u'v dx Therefore we have: xln(x) - ∫x1/x dx = xln(x) - ∫1 dx = xln(x) - x (+c)