Turn on thread page Beta
    • Thread Starter
    Offline

    16
    ReputationRep:
    hey guys,
    just cam back from uni. had a test which is like 15% of my cw. pretty confident but just some question that if you can confirm i got right that'll be great.

    1)  8x\sqrt (x^2 + 1) dx integration by substitution between x=o (lower limit) and x=1 (upper limit). Did anyone get 6?

    2) can;t remember exactly by if you have something like  y= \frac{some function}{2} + 2x , to find dy/dx will you just use the quotient rule to find the first part and then "+ 2"?

    thanks
    Offline

    2
    ReputationRep:
    I could be wrong, but I think:

    1) Answer = (8/3)*2^(3/2). After integrating, I got (8/3)(x^2+1)^(3/2) -- what did you have?

    2) You could use the quotient rule, but it would be a bit pointless. Just differentiate the numerator, then divide it by 2.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Prokaryotic_crap)
    hey guys,
    just cam back from uni. had a test which is like 15% of my cw. pretty confident but just some question that if you can confirm i got right that'll be great.

    1)  8x\sqrt (x^2 + 1) dx integration by substitution between x=o (lower limit) and x=1 (upper limit). Did anyone get 6?

    2) can;t remember exactly by if you have something like  y= \frac{some function}{2} + 2x , to find dy/dx will you just use the quotient rule to find the first part and then "+ 2"?

    thanks
    I'm afraid your first answer is wrong. There is a surd in the answer.

    \frac{8(2\sqrt2-1)}{3}

    There is no need to use the quotient rule on the second question (although you should have got the correct answer even if you did).

    \frac{dy}{dx}=\frac{1}{2}\frac{d  (some function)}{dx}+2
    Offline

    0
    ReputationRep:
    Oops garbage, damn internet lol. Ignore this post please
    Offline

    0
    ReputationRep:
    I got the answer as (16*sqrt(2))/3 - 8/3 for the first one
    • Thread Starter
    Offline

    16
    ReputationRep:
     y = 8x \sqrt(x^2 +1) dx             u= x^2+1  \frac{du}{dx} = 2x    dx= \frac{du}{2x} x= 1  u=2  x=0  u=1

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    8(u -1)^\frac{1}{2} u^\frac{1}{2} \frac{du}{2(u-1)^\frac{1}{2}
    simplyfying that and subbung u=2 and u=1 and then taking the latter from the former will give 6 i'm sure. anyone who disagrees please post working. it'll really help.
    Offline

    0
    ReputationRep:
    u = sqrt(1+x^2), udu/dx = x, dx = u/x * du substitution into the integral and you get 8u^2 du to integrate and your limits become u=1 to u = sqrt(2)
    • Thread Starter
    Offline

    16
    ReputationRep:
    i have just realised my mistake, and to make things worse i have just realised that i had written the right thing and then just crossed it out and written something else. I want to kill myself right now
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    PC,

    This is what it should have looked like.

    \int^1_0 8x \sqrt{x^2 +1} \,dx

    u= x^2+1

    du=2xdx

    4 \int^2_1 u^{\frac{1}{2}} \,du

    I'm sure you can finish it off.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 13, 2008

2,267

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
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.