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    I am having problems doing problems when I am given a u as in a form u^2 = this and that.

    If I were to pick my own value, I'd usually just pick "u", even if it's rooted. But whenever I have to take it squared, or u rooted, I encounter problems.

    Usually I just integrate, and then substitute for dx.

    With u^2's I get really confused.

    For example, in c4, exercise 6H number 4.

    I am given x√(2-x) dx

    There needs to be an EXACT value to be found, and all that has to be integrated with limits from 2 to 0.

    While I did took u as a 2-x, I could only find the approximate value, while when looking at the worked answer, it was suggested I take u^2. I would have never though of that.. and I am struggling in knowing when to take u^2 or rooted u or normal u.

    Because by taking u^2, you'd end up with a surd fraction, and it will also make you change limits with one of them being root 2.


    So yeah.. I am really having troubles with these... any advice from someone who is good or has experienced the same thing?
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    (Original post by studentFlipper)
    I am having problems doing problems when I am given a u as in a form u^2 = this and that.

    If I were to pick my own value, I'd usually just pick "u", even if it's rooted. But whenever I have to take it squared, or u rooted, I encounter problems.

    Usually I just integrate, and then substitute for dx.

    With u^2's I get really confused.

    For example, in c4, exercise 6H number 4.

    I am given x√(2-x) dx

    There needs to be an EXACT value to be found, and all that has to be integrated with limits from 2 to 0.

    While I did took u as a 2-x, I could only find the approximate value, while when looking at the worked answer, it was suggested I take u^2. I would have never though of that.. and I am struggling in knowing when to take u^2 or rooted u or normal u.

    Because by taking u^2, you'd end up with a surd fraction, and it will also make you change limits with one of them being root 2.


    So yeah.. I am really having troubles with these... any advice from someone who is good or has experienced the same thing?
    What was wrong with u = \sqrt{2-x}? It's the same substitution as u^2 = 2-x, and you'll go through the same motions to evaluate the integral. Would you mind posting your working where you try using the u rather than u^2 substitution?
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    (Original post by studentFlipper)
    I am having problems doing problems when I am given a u as in a form u^2 = this and that.

    If I were to pick my own value, I'd usually just pick "u", even if it's rooted. But whenever I have to take it squared, or u rooted, I encounter problems.

    Usually I just integrate, and then substitute for dx.

    With u^2's I get really confused.

    For example, in c4, exercise 6H number 4.

    I am given x√(2-x) dx

    There needs to be an EXACT value to be found, and all that has to be integrated with limits from 2 to 0.

    While I did took u as a 2-x, I could only find the approximate value, while when looking at the worked answer, it was suggested I take u^2. I would have never though of that.. and I am struggling in knowing when to take u^2 or rooted u or normal u.

    Because by taking u^2, you'd end up with a surd fraction, and it will also make you change limits with one of them being root 2.


    So yeah.. I am really having troubles with these... any advice from someone who is good or has experienced the same thing?
    I just did this with u=2-x
    Why would you get an approximate answer?

    The limits change from 0,2 to 2,0 and you have some square roots of u
    So the answer is a multiple of root2
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    (Original post by studentFlipper)
    <snip>
    (To link between my answer and TenOfThem's, usually more than one substitution will work.)
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    (Original post by Smaug123)
    What was wrong with u = \sqrt{2-x}? It's the same substitution as u^2 = 2-x, and you'll go through the same motions to evaluate the integral. Would you mind posting your working where you try using the u rather than u^2 substitution?
    (Original post by TenOfThem)
    I just did this with u=2-x
    Why would you get an approximate answer?

    The limits change from 0,2 to 2,0 and you have some square roots of u
    So the answer is a multiple of root2


    Well for me it's not that easy taking u^2, it does confuse me. I am really struggling with math, and sometimes I spent hours staring at a problem or even at a worked out example without really understanding.

    Anyway, I drew the worked example of taking 2-x as a u, but I probably messed up with signs on the way or broke some rules.
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    (Original post by studentFlipper)
    Well for me it's not that easy taking u^2, it does confuse me. I am really struggling with math, and sometimes I spent hours staring at a problem or even at a worked out example without really understanding.

    Anyway, I drew the worked example of taking 2-x as a u, but I probably messed up with signs on the way or broke some rules.
    I haven't checked your working, but why have you converted everything to decimals?

    You should know that

    2^{3/2} = 2 \times 2^{1/2} = 2\sqrt{2}

    and similarly

    2^{5/2} = 2^2 \times 2^{1/2} = 4\sqrt{2}

    so you can write everything as a multiple of \sqrt{2}

    (also your notation is slightly wrong - you should have written dx = -du not dx = -1 so that your 'u integral' ends with -du instead of -1. Always make it clear what you're integrating with respect to)
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    (Original post by studentFlipper)
    Well for me it's not that easy taking u^2, it does confuse me. I am really struggling with math, and sometimes I spent hours staring at a problem or even at a worked out example without really understanding.

    Anyway, I drew the worked example of taking 2-x as a u, but I probably messed up with signs on the way or broke some rules.
    As I said you have everything in root2s

    Do not change to decimal
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    (Original post by studentFlipper)
    Well for me it's not that easy taking u^2, it does confuse me. I am really struggling with math, and sometimes I spent hours staring at a problem or even at a worked out example without really understanding.

    Anyway, I drew the worked example of taking 2-x as a u, but I probably messed up with signs on the way or broke some rules.
    So the u^2 = 2-x substitution is essentially identical:
    differentiating both sides of that expression by the chain rule, \dfrac{du}{dx} \times 2u = -1, so -\text{d}x = 2u \text{d}u and the integral becomes \displaystyle \int_{\sqrt{2}}^0 (2-u^2)u \times -2u \ \text{d}u.

    This is a polynomial expression of u so it's easy to evaluate.
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    (Original post by Smaug123)
    So the u^2 = 2-x substitution is essentially identical:
    differentiating both sides of that expression by the chain rule, \dfrac{du}{dx} \times 2u = -1, so -\text{d}x = 2u \text{d}u and the integral becomes \displaystyle \int_{\sqrt{2}}^0 (2-u^2)u \times -2u \ \text{d}u.

    This is a polynomial expression of u so it's easy to evaluate.


    Aagghhh it's too complicated for me, I don't understand it. I know the essence of the chain rule,. I just don't see how it is applicable here, I can't get my head around it. I see that you differentiated both sides u^2 >> 2u and 2 - x >>> -1 but why you would multiply 2u by 2u/dx or why do we need a negative dx is what I don't get either. Usually when I am doing integrals like these, I am solving for dx or dt, and then plugging it into the integral. Even considering I do all that, I am still somewhat confused with how we got root 2 in there, this is all so complicated for me when it should be easy.
    (Original post by TenOfThem)
    As I said you have everything in root2s


    Trying to grasp my head around it, too problematic to understand. I have everything in root of 2 as in my u's because they are rooted? You mean like turn 2u^3/2 into something like 2u times *root* u?




    Do not change to decimal
    (Original post by davros)
    I haven't checked your working, but why have you converted everything to decimals?

    You should know that

    2^{3/2} = 2 \times 2^{1/2} = 2\sqrt{2}

    and similarly

    2^{5/2} = 2^2 \times 2^{1/2} = 4\sqrt{2}

    so you can write everything as a multiple of \sqrt{2}

    (also your notation is slightly wrong - you should have written dx = -du not dx = -1 so that your 'u integral' ends with -du instead of -1. Always make it clear what you're integrating with respect to)

    Yes you are right! I should have used 1*(-du) instead. I'll try to remember that, I also know the powers multiplication conversion. The problem for is, that I think I wouldn't have thought of changing them to the roots in the first place, even when asked to give an exact value. I just wouldn't have thought of it. I don't know what I need, maybe mass practice... but as soon as I get to review exercises I forget half the stuff I tried to learn haha
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    (Original post by studentFlipper)
    Aagghhh it's too complicated for me, I don't understand it. I know the essence of the chain rule,. I just don't see how it is applicable here, I can't get my head around it. I see that you differentiated both sides u^2 >> 2u and 2 - x >>> -1 but why you would multiply 2u by 2u/dx or why do we need a negative dx is what I don't get either. Usually when I am doing integrals like these, I am solving for dx or dt, and then plugging it into the integral. Even considering I do all that, I am still somewhat confused with how we got root 2 in there, this is all so complicated for me when it should be easy.
    \dfrac{d}{dx} u^2 = 2 u \dfrac{du}{dx}. That's the chain rule. The negative dx doesn't mean anything - I just lazily didn't multiply through by -1 to give us a \text{d}x = expression.
 
 
 
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