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Homework Help: 3 Basic Integral questions

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Simplify and Integrate:

    1. ∫ √x (3x - 2) dx

    2. ∫ 4x2 - 2 √x / x dx

    3. ∫ (1 - x)3 dx

    The answers are :

    1. 6x5/2 / 5 - 4x3/2 / 3 + C

    2. 2x2 -4 √x + C

    3. x - 3x2 / 2 + x3 - x4 / 4 + C

    2. Relevant equations

    3. The attempt at a solution

    1. x1/2 (3x - 2) dx

    x3/2 / 3/2 * x3 -2x

    2/3 x3/2 * x3 -2x + C

    2. 4x2 - 2x1/2 / x

    (4x - 2 * 1/2) <--- I divided by 'x'

    x4 - 2x + C

    3. 1 - x3

    1x - x4/4

    I can't seem to get the right answers according to the book, could someone explain their process?

    Apologies for writing it from left to right instead of fractional form as you would normally write it on paper, but I haven't been able to write it out that way properly on the forum.
  2. jcsd
  3. Mar 12, 2013 #2


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    It's probably best to put each of these in its own thread.

    There appear to be typos -- that and/or some pretty bad basic algebra.

    Let's look at #1.

    Upon multiplying through by [itex]\displaystyle \ \sqrt{x}\,,\ [/itex] the integrand becomes [itex]\displaystyle \ 3x^{3/2}-2x^{1/2}\ .\ [/itex] What is the anti-derivative of that ?
  4. Mar 12, 2013 #3

    Simon Bridge

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    Not sure I follow your notation in 1 and 2.
    For 3. $$\int (1-x)^3 dx$$ ... they seem to have expanded the brackets first. ##(1-x)^3 \neq 1-x^3##.

    If I have followed your notation correctly - then you have made a similar error in the others. Check how you have multiplied out the brackets.
    Last edited: Mar 12, 2013
  5. Mar 13, 2013 #4
    Looking back at the solution process, it seems everything is written as it should be. For the first one, I never divided through by x, I simply changed the radical sign to a power of 1/2, then used one of the rules where xn+1 / n + 1 + C. I then multiplied the denominator so that it appears beside the whole term and proceeded to get the inverses or anti derivatives of 3x and 2, which would be x3 and 2x respectively.

    As for the third one, I think I did expand the exponent of 3 throughout the brackets, I just got 13 which is 1, and then the x3, then from that I used the xn+1 / n + 1 + C rule to get x4 / 4 and thats where I got confused.
  6. Mar 13, 2013 #5


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    Where did you ever get that x3 ?

  7. Mar 13, 2013 #6


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    Let's look at first two problems.
    Per the instructions, you're supposed to simplify first, and then integrate. In this context that would mean carry out the multiplication. I can't tell what you did here.

    For starters, the way you wrote your integrand, it is 4x2 - 2(√x/x). If this isn't what you meant, then you need to add parentheses around the entire numerator.

    √x / x ≠ 1/2
    Also, in the start of your work, there is no indication that you're integrating, other than the presence of dx.
  8. Mar 13, 2013 #7


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    Put in numbers: is [itex](1-2)^3 = 1^3 - 2^3[/itex]?

    In all your attempts above, your problem is not with calculus, but with algebra. You will never master the first if you don't master the latter.
  9. Mar 13, 2013 #8

    Simon Bridge

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    ##(1+x)^3=(1+x)(1+x)(1+x)=1+3x+3x^2+x^3\neq 1+x^3##
    ... do you see how that is so?
    (Remember when you learned about quadratic equations you learned how to multiply out brackets?)

    The basic rule is: ##a(b+c)=ab+ac##

    It follows that:

    Examples:$$\int x(x^2+2)dx = \int (x^3+2x)dx = \frac{1}{4}x^4+x^2 + c$$
    $$\int \frac{2x^3-4}{x^2}dx = \int \frac{1}{x^2}\left ( 2x^3 - 4 \right )dx = \int (2x-4x^{-2})dx = x^2 +2x^{-1} + c$$

    After that there are a few basic concepts (like ##x=\sqrt{x}\sqrt{x}##) - I think you should pick one and go through it step by step explaining your reasoning on each line.
    Last edited: Mar 13, 2013
  10. Mar 13, 2013 #9

    The x3 I got from taking the integral of 3x, which now that I think about it might be wrong.


    Well, what I did there was remove the radical, and replace it with its equivalent 1/2 power, then I used that rule of xn+1 / n + 1 + C to get 3/2.

    The second problem I did indeed forget to add parenthesis for the numerator, it should look like this:

    (4x2 -2 √x) / x

    You're saying to expand the x1/2 into the brackets? I'm not entirely sure how to do that, this is where my Algebra fails me and has done so for many years. Regardless of how many books I read, or how many Stanford math videos or MIT math videos I've watched (Algebra related), I still haven't become comfortable with Algebra.


    Would I be a terrible person if I said thats exactly what I thought it would equal?

    I'm honestly not sure what to do anymore. I've spent far too much time going back into a subject like algebra only to come out of it with nothing...again. I'd like to continue with integration, however, I'll just have the same problems time and time again.

    Whilst writing this, Simon Bridge had written his post, that's why everything took so long.

    Simon Bridge,

    I'm not going to lie, it took me quite a while to figure out how you managed to get 1 + 3x + 3x2 + x3 as an answer. As for the first example , I can follow that, but the second one I was confused with how you distributed your 1/x2 into the bracketed term. 1/x2 * 2x3 = 2x? And then you get a negative exponent on the 4x? How did that come about?
  11. Mar 13, 2013 #10


    Staff: Mentor

    Yes, it's wrong.
    You have to do the multiplication first. IOW you have to multiply x1/2(3x - 2). You can't just integrate the factors separately.
    If you want to be successful in calculus, you have to become competent with algebra (and trig). There's no way around it. It's not enough just to read the books or view the videos - you have to work the problems.
    IMO, you're wasting your time if you don't make an effort to get competent with the basics. You might be able to understand some of the high-level concepts of calculus, but if you don't have a good grasp of the basic foundation, you won't be able to work a problem all the way through, or even be able to follow someone else's work.

  12. Mar 13, 2013 #11

    Simon Bridge

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    Well - I may have made a mistake, that can happen when trying to demonstrate something on the fly... lets have a look in more detail... the problematic example was:
    You wanted to know how the second step turned into the third one? Here it is, step-by-step, without the integral signs to confuse things:$$\frac{1}{x^2}\left ( 2x^3 - 4 \right ) = \frac{1}{x^2}(2x^3) + \frac{1}{x^2}(-4) = \frac{2x^3}{x^2}-\frac{4}{x^2}=2x - 4x^{-2}$$... that was just ##a(b+c)=ab+ac## where ##a=1/x^2##, ##b=x^3##, and ##c=-4##.

    The negative exponent is because $$a^{-n}=\frac{1}{a^n}$$... so I could type it out quickly. It's also useful for applying integration rules since all exponents apart from "-1" follow the same rule.

    It looks like I did make a mistake with the integration though. $$\int (2x-4x^{-2})dx=2\int xdx-4\int x^{-2}dx = x^2+4x^{-1}+c$$ since: $$\frac{d}{dx} \left ( x^{-1} + c \right ) = -x^{-2}\\ \Rightarrow x^{-1} + c = -\int x^{-2}dx$$ (... to see how I did that - integrate both sides in the top line.)
    Yes they will - time and time again. Algebra is to math what grammar is to language - you would not expect to do well with literature without first understanding grammar would you?

    Your algebra troubles are very serious and you will not advance in calculus without addressing them. They are so severe that for us to continue coaching you here is likely a waste of your time and ours.

    To continue with calculus you need an algebra tutor - someone who is prepared to sit down and watch you work through the problems and patiently show you where your thinking is out.
  13. Mar 13, 2013 #12

    Simon Bridge

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    Reviewing - I am concerned that I may not have been as clear as I needed to be before...
    The fraction sign is a division - not just a fancy way to write a number.
    So: $$\frac{a}{b}=a\div b$$ Multiplying fractions goes like this: $$\frac{a}{b}\frac{c}{d}=\frac{ac}{bd}$$ but we have to bear in mind that every number is an implied fraction so: $$a \Leftrightarrow \frac{a}{1}$$ so $$\frac{1}{a}b=\frac{1}{a}\frac{b}{1}=\frac{b}{a}=b\div a$$... applying that to the example (last post) $$\frac{1}{x^2}(2x^3)=\frac{1}{x^2}\frac{2x^3}{1}=\frac{2x^3}{x^2}=2x^3\div x^2 = 2x$$ ...but we'd normally think of it like this: $$ \frac{2x^3}{x^2}=\frac{2x(x^2)}{x^2}=\frac{2x}{1} \frac{x^2}{x^2}=\frac{2x}{1}\frac{1}{1}=2x$$ ... where we notice that ##2x^3=2x\times x^2## and apply the multiplication rule (earlier) in reverse.


    Aside: there's a caveat here - ##2\frac{1}{2}## is read as "two and a half" in English text, but it is read "two multiplied by one divided by two" (i.e. "one") in algebra.
  14. Mar 14, 2013 #13

    Alright, thanks for the clarity. And I agree, I would need to go over Algebra again, perhaps more vigorously than before.

    Simon Bridge,

    Ah, I see, that step-by-step of the example helped out. For some reason I didnt see the 1 / x2 as what it really is, its difficult to explain. I was trying to figure out how to multiply 1 / x2 by 2x3. If I saw .5 instead of a 1 / 2, I could instantly tell that .5 * 2 would be half of two. When I saw 1 / x2 I was trying to see how it would go into 2x3, I find it hard to explain what I was thinking, sorry.

    As for your second post, the worst part is, I remember the multiplication of fractions rule, as well as the fact that every number is an implied fraction. I know that I'm mixing up my rules now because when I saw your explanation of fraction multiplying, I considered to cross multiply for some reason...

    Anyway, I know I need to return to Algebra basics, and perhaps a tutor is that final solution, otherwise I really dont know what else could be done. Thanks a lot for the detailed explanations, I can never get enough of those.

    I'm still watching some math foundation videos as recommended by this forum, but Id also like to know if you personally had seen or heard of any good books that could help me, seeing as how much I benefit from explanations, perhaps there is a book you know of regarding Basic/Intermediate/Advanced algebra? I dont want to say I'm looking for a dumbed-down version, but maybe thats something that will actually help? I dont know.
  15. Mar 14, 2013 #14

    Simon Bridge

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    The "cross-multiply" thing you vaguely remember is probably the common way of teaching the addition of fractions... $$\frac{a}{b}+\frac{c}{d} = \frac{ad+cb}{bd}$$ ... actual "cross multiply" has a different meaning in algebra.

    You should be able to work out the various rules from the multiplication ones already given above... i.e. $$\frac{a}{c}+\frac{b}{c} = \frac{1}{c}(a+b) = \frac{a+b}{c}$$ ... which you've seen before written in the other order.

    If the numerators are different: $$\frac{a}{b}+\frac{c}{d}$$... you need to make them the same. You do this by multiplying each term by 1 like this: $$\frac{a}{b}\frac{x}{x}+\frac{c}{d}\frac{y}{y}$$ ... see how that doesn't change the result?
    You have to pick your x and y so that bx=dy ... which means that the denominators are the same. There's an infinite number of possibilities to choose from, and some choices are more helpful than others, so you may as well pick the one that makes the math easy. If you play around with it you'll see that the general case is x=d and y=b ... but there are shortcuts where b is a whole multiple of d etc.

    Subtraction is the same as addition by a negative number.
    Division is the same as multiplication by a fraction.
    That's pretty much all of arithmetic.

    Algebra manipulating the form that an expression takes to make some calculation easier. At your level that means multiplying out brackets - maybe implied brackets.

    i.e. It is easier to see how to integrate ##x^2+2x+1## than ##(x+1)^2## even though they are the same thing. So - instead of doing what's in the brackets first like you learned in "order-of-operations" lessons, it actually makes life easier to multiply out the brackets first... this time.

    Books and videos are not it... though there is an Algebra for Dummies (nothing wrong with needing help - it's not asking for it that's stupid) what you have to do is lots and lots of examples.
  16. Mar 14, 2013 #15
    Well, surprisingly enough, I never read that book, I dont know why. Nevertheless, I'll take a look at it.

    And yes, those examples and methods are reminiscent of those videos I've been watching. They may not be strictly algebra related, however the videos do contain what I think are the basics and essentials.


    Thanks again for the reply, I'll probably be back soon.
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