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AP.CALC:
FUN‑6 (EU)
,
FUN‑6.C (LO)
,
FUN‑6.C.1 (EK)
,
FUN‑6.C.2 (EK)

Video transcript

let's say that we wanted to take the indefinite integral of x squared times 3x minus 1 DX pause this video and see if you can evaluate this so you might be saying oh what what kind of fancy technique could I use but you will see sometimes the fanciest or maybe the least fancy but the best technique is to just simplify this algebraically so in this situation what will happens if we distribute this x squared well then we're going to get a polynomial here within the integral so this is going to be equal to the integral of x squared times 3x is 3x to the third and then negative 1 times x squared is minus x squared and then that times DX and now this is pretty straightforward to evaluate this is going to be equal to the antiderivative of X to the third is X to the fourth over 4 so this is going to be 3 times X to the fourth over 4 I could write it that way or let me just write it X to the fourth over 4 and then the antiderivative of x squared is X to the third over 3 so minus X to the third over 3 this is an indefinite integral there might be a constant there so let me write that down and we're done the big takeaway is you just have to do a little bit of distribution to get a form where it's easy to evaluate the antiderivative let's do another example let's say that we want to take the indefinite integral of there's going to be a hairy expression so X to the 3rd plus 3x squared minus 5 all of that over x squared DX what would this be pause the video again and see if you can figure it out so once again your brain might want to try to do some fancy tricks or whatever else but the main insight here is to realize that you could just simplify it algebraically what happens if you just divide each of these terms by x squared well then this thing is going to be to put some parentheses here X to the third divided by x squared is just going to be X 3x squared divided by x squared is just three and then negative five divided by x squared you could just write that as negative five times X to the negative two power and so once again we could just need to use the reverse power rule here to take the antiderivative this is going to be let's see the antiderivative of X is x squared over two x squared over two plus the antiderivative of three would just be three X the antiderivative of negative five x to the negative two so we would increment the exponent by one positive one and then divide by that value so it would be negative five x to the negative one we're adding one to negative one all of that divided by negative one which is the same we could write it like that well this this these two would just you'd have minus and you're dividing by negative one so it's really just going it can rewrite it like this plus five x to the next one and you could take the derivative of this to verify that it would indeed give you that and of course we can't forget our plus C never forget that if you're taking it if you're digging an indefinite integral all right let's just do one more for good measure let's say we're taking the indefinite integral of the cube root of x to the fifth DX pause the video and see if you can evaluate this try to write a little bit neater X to the fifth DX pause the video and try to figure it out so here the realization is well if you just rewrite all this as one exponent so this is equal to the indefinite integral of X to the fifth to the one-third I just rewrote the cube root as the one-third power DX which is the same thing as the integral of x to the if I raise something to a power and then raise that to a power I can multiply those two exponents that's just exploited properties X to the 5/3 DX many of you might have just gone straight this step right over here and then once again we just have to use the reverse power rule this is going to be X to the we increment this 5/3 by 1 or we can add 3/3 to it so it's X to the 8 thirds and then we divide by 8 thirds or multiply by its reciprocal so we could just say 3/8 times X to the 8th thirds and of course we have our plus C and verify this you use a power rule here you'd have 8/3 times with 3/8 we just give you coefficient of 1 and then you decrement this by 3/3 or 1 you get to 5/3 which is exactly what we originally had so the big takeaway of this video many times the most powerful integration technique is literally just algebraic simplification first
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