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if depreciationamortisation is done properly impairment adjustments will not arise requireddo you agree with the above statement critically and
method of cylinders or method of shellsthe formula for the area in all of the cases will be
volumes of solids of revolution method of cylindersin the previous section we started looking at determine volumes of solids of revolution in
formulas for the volume of this solidv intba a x dx v intdc a y dywhere a x amp a y is the
some important issue of graphbefore moving on to the next example there are some important things to notefirstly in almost all problems a graph is
the mean value theorem for integralsif f x is a continuous function on ab then there is a number c in ab such
average function valuethe first application of integrals which well see is the average value of a function the given fact tells us how to calculate
integrate followingint -2 24x 4- x2 1dxsolutionin this case the integrand is even amp the interval is accurate soint -2 24x 4- x2
even and odd functions this is the final topic that we have to discuss in this chapter firstly an even function is any function which
evaluate followingint 0ln 1 pi excos1-exdxsolutionthe limits are little unusual in this case however that will happen sometimes therefore dont
there are really three various methods for doing such integralmethod 1this method uses a trig formula as intsinx cosx dx frac12 intsin2x dx -14
constants of integrationunder this section we require to address a couple of sections about the constant of integration during most calculus class we
types of infinity mostly the students have run across infinity at several points in previous time to a calculus class though when they have dealt
fundamental theorem of calculus part ii assume fx is a continuous function on ab and also assume that fx is any anti- derivative for fx henceaintb
fundamental theorem of calculus part iif fx is continuous on ab sogx aintx ft dtis continuous on ab and this is differentiable on a b and asgprimex
proof of if fx gt gx for a lt x lt b then aintb fx dx gt gxbecause we get fx ge gx then we knows that fx - gx ge 0 on a le x le b and therefore
proof of int fx gx dx int fx dx intgx dxit is also a very easy proof assume that fx is an anti-derivative of fx and that gx is an anti-derivative
proof of various integral factsformulaspropertiesin this section weve found the proof of several of the properties we saw in the integrals section
rolles theorem assume fx is a function which satisfies all of the following1 fx is continuous in the closed interval ab2 fx is differentiable in
fermats theorem if fx has a relative extrema at x c and fprimec exists then x c is a critical point of fx actually this will be a critical point
proof of limqrarr0 cosq -1q 0we will begin by doing the followinglimqrarr0 cosq -1q limqrarr0cosq - 1cosq 1q cosq 1 limqrarr0cos2q - 1 q cosq
chain rule if fx and gx are both differentiable functions and we describe fx f gx so the derivative of fx is fprimex f primegx
quotient rule fg fg - fgg2here we can do this by using the definition of the derivative or along with logarithmic definitionproof here we do the
product rule f gprime f prime g f gprimeas with above the power rule so the product rule can be proved either through using the definition of the
power rule dxndx nxn-1there are really three proofs which we can provide here and we are going to suffer all three here therefore you can notice all