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Math Help - more basic calculus

  1. #1
    Junior Member turbod15b's Avatar
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    more basic calculus

    can some show me step by step please
    thanks

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  2. #2
    No one in Particular VonNemo19's Avatar
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    Show you what. I don' see anything. Are you talking about how you want someone to show you "Step-by-Step" the sitcom? I hated that show.

    Nevermind, I can see it now.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    f) \frac{1}{4}x^4+C The only step you need for this is the power rule. Do you know the power rule? Oh and algebra too. dy=x^3dx\Longleftrightarrow\frac{dy}{dx}=x^3
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Use the power rule for g too...........

    \frac{1}{3}x^3+x^2-3x+C

    You do know the power rule right?
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  5. #5
    Junior Member turbod15b's Avatar
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    thanks
    but can u show me how to do it for all of them
    thanks
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  6. #6
    Newbie darthfader's Avatar
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    j) Let me do j for you step by step.
    \int{\frac{{x}}{{x^2  + 2}}}dx

    Let u = x^2 + 2
    Then du = 2xdx

    Notice that our du is the same as the numerator of our given except of the numerical coefficient 2. Do you see it?

    Now, we can put the 2 by doing this.

    \frac{1}{2}\int{\frac{2xdx}{x^2+2}} Is this equal to the original given? Of course yes. So we can continue.

    Then
    \frac{1}{2}\int{\frac{2xdx}{x^2+2}} = \frac{1}{2}\int{\frac{du}{u}}.

    There is a formula in integral calculus that states that the \int{\frac{du}{u}} = \ln\left|u\right| + C.

    Hence in our problem..

    \frac{1}{2}\int{\frac{2xdx}{x^2+2}} = \frac{1}{2}\int{\frac{du}{u}} = \frac{1}{2}\ln\left|u\right| + C.

    Plugging in u.

    \int{\frac{{x}}{{x^2  + 2}}}dx = \frac{1}{2}\ln\left|x^2 + 2\right| + C

    You can check by getting the derivative of our final answer. If it results to the given, then we're pretty much correct.

    I hope this helps.

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  7. #7
    Junior Member turbod15b's Avatar
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    thanks bro
    can u help me to the other 1s aswell
    cheers
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  8. #8
    Newbie darthfader's Avatar
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    For h, let me ask first, is it a Definite Integral? With limits 2 and 3? Do you know how to evaluate Definite Integrals?

    The problem is quite easy and you can try it by yourself.

    Try to continue this, just feel free to ask if you have some questions.

    Hint: Start with this:

    2\int_2^3 {d\theta  + } \frac{1}{2}\int_2^3 {\cos 2\theta (2d\theta )}

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  9. #9
    Junior Member turbod15b's Avatar
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    im lost
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  10. #10
    Newbie darthfader's Avatar
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    Can you tell me what part of solving the problem you find difficulties?
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  11. #11
    No one in Particular VonNemo19's Avatar
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    Hey darth! If I were you, I would change my signature. You should write the work function so that it doesn't make you look like you don't know that the limit of a constant is that constant.

    I'm just joking with you. Sometimes people can't pick up on humor through text.

    The limit as x tends to infinity of work is, guess what?

    MORE WORK! Get it?
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  12. #12
    Newbie darthfader's Avatar
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    Thanks. I'll think about changing my signature. Maybe I can come up with a better one. Haha.
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