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To solve this we must use integration by parts: int(udv) = uv - int(vdu) (1) Hence let u = ln(x), dv = dx => du=(1/x)dx, v=x, and now using (1) and substituting values we obtain int(ln(x)dx) = ln(x)x -...
Integration can be viewed in many ways. The most common way to interpret an integral is to take the area under the curve you would like to integrate. For example Draw y=x, limit between 0 and 1, shade...
To integrate this we need to use the chain rule, substituting cos2x = u Integral becomes: u2sin32x dxChain rule: dy/dx = du/dx dy/du du/dx = -2sin2x --> dx = -1/2sin2x du Substitu...
Differentiation is about the tangent at a point, from GCSE we have a formula to work out the gradient between two points, in other words, the gradient of the tangent. This means that as the two points get...
(a) We can ‘get rid of’ a square root in the denominator simply by multiplying by 1 (value of the fraction stays unchanged) in a suitable form. We will take advantage of this formula: (a+b)(a-b)=a^2...
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