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    Hi,

    How do you do ∫ 2^x dx? And what is the answer?

    Thanks!
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    Let y = a^x
    Taking logs of both sides,
    ln y = ln a^x
    ln y = x ln a
    Differentiating (using chain rule on left hand side),
    1/y * dy/dx = ln a (because x ln a is just a constant multiple of x, like 3x)
    But y = a^x.
    1 / a^x * dy/dx = ln a
    dy/dx = a^x ln a.

    So if you differentiate 2^x, you'll get 2^x ln 2. Integration is the opposite of differentiation, and so integrating 2^x you'll get (2^x) / ln2 (+ C).
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    I know that if you differentiate 2^x you get 2^xln2, but I can't see how integrating 2^x gives (2^x) / ln2 (+ C). Is it that to differentiate it you times by ln2 so to integrate it you divide by ln2? Can you do that?
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    (Original post by sweet_gurl)
    I know that if you differentiate 2^x you get 2^xln2, but I can't see how integrating 2^x gives (2^x) / ln2 (+ C). Is it that to differentiate it you times by ln2 so to integrate it you divide by ln2? Can you do that?
    Yeah, more or less. Say you had a function y of x such that dy/dx = 2^x. It's obvious that y = k.2^x (some constant multiple), because to differentiate you just multiply by ln 2.

    d/dx k.2^x = k d/dx 2^x = k . ln2 . 2^x
    but we said that dy/dx = 2^x.
    k . ln2 . 2^x = 2^x
    k = 1/ln2
    y = 2^x / ln2.

    It's just using the fact that integration and differentiation are opposites. d/dx e^2x = 2e^2x, therefore ∫ e^2x dx = 1/2 e^2x.
    • Thread Starter
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    Ahh, I see. Thank you very much for your help!
    • Thread Starter
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    I've got another question:

    How would you integrate (sinx)^3 dx ?
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    You could use (If you do Further Maths)

    cos^3x = [1/2(z+1/z)]^3
    =>cos^3x = 1/8[z^3 + 3z + 3/z + 1/z^3]
    =>cos^3x = 1/8[(z^3 + 1/z^3) +3(z+1/z)]
    =>cos^3x = 1/8[2cos3x + 6cosx]

    which you can integrate easily.
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    split sin^3x into sinx(sin^2x)

    rearrange sin^2x into 1-cos^2x

    Mulitply out and integrate

    In the spoiler is a hint if you get stuck.

    Spoiler:
    Show
    remember \tex \large \int sinxcos^nx = -\frac{1}{n+1}cos^x + c
    • Thread Starter
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    Thanks! My answer includes a (cosx)^3 term. Does this matter, or do you have to simplify it further?
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    (Original post by sweet_gurl)
    Thanks! My answer includes a (cosx)^3 term. Does this matter, or do you have to simplify it further?
    I think cos3x is fairly simple. Do you mean do you have to convert it to sinx? Nah.
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    If you want another way to integrate a^{x} use a = e^{\ln a} so

    \int a^{x}dx = \int (e^{\ln a})^{x} dx = \int e^{x\ln a} dx

    Then you just use \int e^{kx} dx = \frac{1}{k}e^{kx}. Saves having to do the whole differentiation thing.
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    Yep thats how i would do it.
 
 
 

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