cu6l3-integration-by-parts-antidifferentiation

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Good day students in this group of very going over some examples on how to umm, antidifferentiate by parts. So let's go ahead and take a look at the formula for of integration or anti-differentiation by parts per second so the formula is as follows formula so the the in definite integral of u DV is equal to use the minus the integral of the DU okay so this is the formula for him integration by parts of you basically have to select the you and DV and then generates the side of the equation of evaluate the integral of vdu and that will be answer Now how did this formula, about the am answer is the product role this is a product rule of them backwards or the reverse of the product rule okay so as a real quick derivation of derivation of the integration by parts formula. So start with the of product rule so let's see one of find that the town derivative of you the sense of you using the of UV prime notation the use the denotation so D the derivative of UV is equal to the times the derivative of you the new plus you times the derivative of the DV okay now of what if I integrate both sides of this equation. We know we can use term right term integration on the right side. So if I find anti-derivative here and of this to expressions on the right on the left will have is some cancellation action happening this integral is to be cancels out the left with UV equals the integral of the DU plus the integral of you DV okay now what happens if I isolate you DV by subtracting the integral of vdu from both sides in using the reflexive property of equality all have the integral of udv equals UV and I subtract the integral of vdu from both sides minus the integral of vdu right so there goes your of integration by parts formula is a clearly see that it is of a results of integrating the product rule so of one selection of you the most important steps integration by parts is selecting the u okay so to select u selecting u let's write that down selectingu were going to use the LiPET rule okay so that's the rule for selecting you so use LIPET rule so what on earth is the light petrol result is the order for selecting him functions okay so LIPET rule of L is for logarithmic functions so logarithmic and I is for inverse trigonometric so the I for the LIPET is for inverse trigonometric the key is for your polynomial function that he is for exponential and then the T is for trigonometric given the metric right so when you want to select you whichever of of this graph the function should look first that is what you select as your you okay so that's the X of of acronyms like that you want to keep in mind for selecting your you rise take a look at some examples task is to evaluate the following question number one, one of find the definite integral of the cube root of X log base four of X the X okay so this is currently of an integration by parts problem so we see the product of two functions here the cubic function the logarithmic function so the question is which is the you of so we want to have a following the formats which is this right the formula down here so we can keep track of what we doing all I get is the integral of of udv equals UV minus the integral of the DU right so we want to look for which function is you and then the other function, with the X of udv okay so now the use LIPET rule of to see which one comes first so remember light petrol if you have a logarithmic function does the highest priority function so we have a logarithmic here so that automatically our you write so I'll you is going to be log base four of X the X that of makes the other function of DV okay slight error here is six that a new is log base four of X and then DV is going to be of let's try this is X to the one third the X okay pirate this Kubrick as using of the effort property of exponents not only have these two we know what you end DVR now we need of each V and DU okay so how the find the you would basically differentiate of both this equation both sides season differentials. So if I differentiate this of site right here have DU derivative of use is one so DU

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