Hey there! Sign in to join this conversationNew here? Join for free

Can someone go through this question with me step by step? U Substitution watch

    • Thread Starter
    Offline

    22
    ReputationRep:
    Name:  jee.png
Views: 132
Size:  6.7 KB

    I am trying and failing to wrap my head around this question ugh
    Offline

    19
    ReputationRep:
    (Original post by Dinasaurus)
    Name:  jee.png
Views: 132
Size:  6.7 KB

    I am trying and failing to wrap my head around this question ugh
    Find dx/du. Change the limits of integration. Write out the integral with the new limits and with dx/du du in place of dx, not putting things in terms of u right away, and see if anything cancels nicely. Then if you've done it right you need a simple "creative adding of zero" so to speak.
    lol no you don't you just do the fractions normally, I am very tired
    Offline

    22
    ReputationRep:
    (Original post by Dinasaurus)
    Name:  jee.png
Views: 132
Size:  6.7 KB

    I am trying and failing to wrap my head around this question ugh
    If u = 3-x^3 then \frac{du}{dx} = -3x^2 \Rightarrow \frac{dx}{du} = -\frac{1}{3x^2}

    Then your integral is: \displaystyle \int \frac{x^5}{3-x^2} \, \mathrm{d}x =  \int \frac{x^5}{3-x^2} \, \frac{dx}{du}\mathrm{d}u

    You can do that because you can pretend that the du's cancel so: \frac{dx}{du}du = dx.


    So, plugging in the relevant bits: \displaystyle \int \frac{x^5}{u} \times \frac{-1}{3x^2} \, \mathrm{d}u

    Then notice that \frac{x^5}{-3x^2} = -\frac{x^3}{3}

    and that u = 3-x^3 \Rightarrow x^3 = 3-u^2.
    • Thread Starter
    Offline

    22
    ReputationRep:
    (Original post by 13 1 20 8 42)
    Find dx/du. Change the limits of integration. Write out the integral with the new limits and with dx/du du in place of dx, not putting things in terms of u right away, and see if anything cancels nicely. Then if you've done it right you need a simple "creative adding of zero" so to speak.
    lol no you don't you just do the fractions normally, I am very tired
    I found dx as 1/-3x^2 du

    The main issue I have is the fraction, I don't know how to substitute the x^5 on the numerator.

    -u+3 = x^3

    Should I multiply both parts by the power of 5/3?
    Offline

    10
    ReputationRep:
    (Original post by Zacken)
    .
    Use the subsitution  u = 3 - x^3....
    Offline

    22
    ReputationRep:
    (Original post by 16Characters....)
    Use the subsitution  u = 3 - x^3....
    I must be going blind! :lol: Thanks, I think I've edited it and it should be fixed. Embarrassing much...
    Offline

    19
    ReputationRep:
    (Original post by Dinasaurus)
    I found dx as 1/-3x^2 du

    The main issue I have is the fraction, I don't know how to substitute the x^5 on the numerator.

    -u+3 = x^3

    Should I multiply both parts by the power of 5/3?
    No, it cancels in the integral with the dx/du to give you something you can express more easily
    • Thread Starter
    Offline

    22
    ReputationRep:
    (Original post by 13 1 20 8 42)
    No, it cancels in the integral with the dx/du to give you something you can express more easily
    I got down to u-3/3u

    Damn that looks a lot more doable.
    Offline

    19
    ReputationRep:
    (Original post by Dinasaurus)
    I got down to u-3/3u

    Damn that looks a lot more doable.
    think that's right
    Offline

    19
    ReputationRep:
    (Original post by 13 1 20 8 42)
    think that's right

    am I too late?
    Offline

    19
    ReputationRep:
    (Original post by TeeEm)
    am I too late?
    It would appear so
    Offline

    19
    ReputationRep:
    (Original post by 13 1 20 8 42)
    It would appear so
    my new catchphrase ...
    Offline

    21
    ReputationRep:
    Ooh looks fun, but IM LATE IM LATE FOR A VERY IMPORTANT DATE!
    😏
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.