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# Definite integral involving natural log

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.B (LO)
,
FUN‑6.C (LO)

## Video transcript

let's now take the definite integral from two to four of six plus x squared over X to the third power DX and first this might seem pretty daunting I have this rational expression but if we just rewrite this it might jump out at you how this could be a little bit simpler so this is equal to the integral from 2 to 4 of 6 over X to the 3rd power plus x squared over X to the 3rd power DX I just separated this numerator out I just divided each of those terms by X to the 3rd power and this I could rewrite this is equal to the integral from 2 to 4 of 6 X to the negative 3 power that's that first term there and x squared divided by X to the 3rd well that is going to be 1 over X so plus 1 over X DX now this is going to be equal to let's take the antiderivative of the different parts and we're going to evaluate that at 4 and we're going to evaluate that at 2 and we're going to find the difference between this express the antiderivative valuated at 4 and a 2 now what is the antiderivative of 6 X to the negative 3 well here once again we can just use we could use as a power rule for taking the antiderivative or it's the reverse of the derivative power rule we know that if we if we're taking the integral of x to the n DX the antiderivative of that is going to be X to the n plus 1 over n plus 1 and if we were just taking an indefinite integral there would be some plus C the reason why we don't put the plus C's here is when you when you evaluated both both bounds of integration the C would cancel out regardless of what it is so we don't really think about the C much when we're taking definite integrals but let's apply that to 6 X to the negative third power so it's going to be we're going to take X to the negative 3 plus 1 so it's X to the negative 2 and so we're going to divide by negative 2 as well of course we had that six out front from the get-go so that's the antiderivative 6x to the negative 3 power and what's the antiderivative of 1 over X you might be tempted to use the same idea right over here you might be tempted to say alright well the antiderivative of X to the negative 1 which is the same thing as 1 over X would be equal to X to the negative 1 plus 1 over negative 1 plus 1 but what is negative 1 plus 1 it is 0 so this doesn't fit this property right over here but lucky for us there is another property and we went the other way when we were first taking derivatives of natural log functions the antiderivative of 1 over X or X to the negative 1 is equal to sometimes you'll see it written as natural log of X plus C and sometimes and I actually prefer this one because you could actually evaluate it for negative values is to say the absolute value the natural log of the absolute value of X and this is useful because this is defined for negative values not just positive values the negative the natural log of X is only defined for positive values of X when you take the absolute value now it could be negative or positive values of X and it works the derivative of this is indeed 1 over X now it's not so relevant here because our bounds of integration are both positive but if both of our bounds of integration were negative you could still do this by just reminding yourself that this is the natural log of absolute value of X so this we could say is plus the natural log of the absolute value of X it's not a bad habit to do it and if everything is positive well the absolute value of X is equal to X and so what is this going to be equal to this is equal to let's evaluate everything at 4 and actually before I even evaluate at 4 what's 6 divided by negative 2 that's negative 3 so if we evaluated at 4 it's going to be negative 3 over 4 squared 4 to the negative 2 is 1 over 4 squared and then plus the natural log of the we could say the absolute value of 4 but the absolute value of 4 is just 4 the abbe the natural log of four and from that we're going to subtract everything evaluated at two so let's do that so if we evaluated it - it's going to be negative three over two squared so 2 to the negative 2 is 1 over 2 squared over 2 squared plus the natural log of the absolute value of positive 2 is once again it's just 2 and so what does this give us so let's try to simplify it a little bit so this is negative 3/16 we do that same color so this is going to be equal to negative negative 3 sorry not negative 3/16 it's got to be very careful sorry yes sorry it isn't hatest 3/16 for some reason my brain start thinking 4 to the 3rd power negative 3/16 plus natural log of 4 and then this right over here is negative 3/4 negative 3/4 to do that same color this right over here is negative 3/4 we have this negative sign out front that we're going to have to distribute so the negative of negative 3/4 is plus 3/4 plus 3/4 and then we're going to subtract remember we're distributing this negative sign the natural log the natural log of 2 and what is this equal to alright so this is going to be equal to and I'm now going to switch to a neutral color so let's add these two terms that don't involve the natural log and let's see if we have a common number 3 over 4 is the same thing that is the same thing as we multiply the numerator and denominator by 4 that is 12 over 16 and so you have negative 3/16 negative 3 16 plus 12 16 will give you nine sixteenths 9/16 and then we're going to have the ones that do involve the natural log natural log of 4 minus the natural log of 2 so we could write this plus the natural log of 4 - the natural log of - and you might remember from your logarithm properties that this over here this is the same thing as the natural log of 4 divided by 2 this comes straight out of your logarithm properties and so this is going to be the natural log of 2 natural log of 2 so we deserve a little bit of a drum roll now this is all going to be equal to this is going to be equal to the natural log the sr9 over 16 plus the natural log of 2 plus the natural log of 2 and and we are done
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