If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:5:33

𝘶-substitution: multiplying by a constant

FUN‑6 (EU)
FUN‑6.D (LO)
FUN‑6.D.1 (EK)

Video transcript

let's take the indefinite integral of the square root of 7x plus 9 DX so my first question to you is is this going to be a good case for you substitution when you look here we may be the natural thing to set to be equal to U is 7 X plus 9 but do I see its derivative anywhere over here well let's see if we set u to be equal to 7x plus 9 what is the derivative of U with respect to X going to be derivative of U with respect to X is just going to be equal to 7 derivative of 7x is 7 derivative of 9 is 0 so do we see a 7 lying around anywhere over here well we don't but what could we do in order to have a 7 lying around but not change the value of the integral well the neat thing and we've seen this multiple times is when you're evaluating integrals scalars can go in and outside of the integral very easily just to remind ourselves if I have the integral of let's say some scalar a times f of X DX DX this is the same thing as a times the integral of f of X DX that the integral of the scalar times the function is equal to the scalar times the integral of the function so let me put this aside right over here so with that in mind can we multiply and divide by something that will have a 7 showing up well we can multiply and divide by 7 so imagine doing this let's rewrite our original integral the original so let me draw a little arrow here just to go around that aside we could write our original integral as being equal to the integral of 1/7 times 7 times the square root of 7x plus 9 DX and if we want to we could take the 1/7 outside of the integral we don't have to but we could rewrite this as 1/7 times the integral of 7 times the square root of 7x plus 9 DX so now if we set u equal 7x plus 9 do we have its derivative laying around well sure the 7 is right over here we know we know that D u if we want to write in differential form D U is equal to 7 times DX so D U is equal to 7 times DX that part right over there is equal to D U and if we want to care about you well that's just going to be the 7x plus 9 that is our u so let's rewrite this indefinite integral in terms of U it's going to be equal to it's going to be equal to 1/7 times the integral of and I'll just take the 7 and put it in the back so we could just write the square root of u square root of u D u 7 times DX is d u and we can rewrite this if we want as u to the one-half power it makes it a little bit easier for us to kind of do the reverse power rule here so we can rewrite this as equal to 1/7 times the integral of U to the one-half power D U and let me just make it clear this u I could have written in white if I want the same color and this D U is the same d u right over here so what is the antiderivative of U to the one half power well we increment to use power by one so this is going to be equal to and let me not forget this 1/7 out front so it's going to be 1/7 times 1/7 times if we increment the power year it's going to be U to the three halves one half plus one is one and a half or three halves so it's going to be u to the three-halves u to the three-halves and then we're going to multiply the the this new thing times the reciprocal of three halves which is 2/3 and I encourage you to verify the derivative of 2/3 u to the three-halves is indeed u to the one-half and so we have that and since we're multiplying 1/7 times this entire indefinite integral we could also throw in a plus C right over here there might have been a constant and if we want we can distribute the 1/7 so it would get 1/7 times 2/3 is 2 over 21 2 over 21 you to the three-halves and 1/7 times some constant well that's just going to be some constant and so I could write a constant like that I can call that C 1 and then I could call this C 2 but it really is just some arbitrary constant and we're done oh actually no we aren't done we still just have our entire thing in terms of U so now let's unsubstituted so this is going to be equal to 2 over 21 times u to the three-halves and we already know it U is equal to U is equal to 7x plus 9 we put a new color in here just to ease the monotony so it's going to be 2 over 21 times 7x plus 9 to the three-halves power to the 3 halves power plus C plus C and we are done we were able to take a kind of hairy looking integral and realize that even though it wasn't completely obvious at first that u substitution is applicable