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Math Help - integrals

  1. #1
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    integrals

    Hi everyone,

    Could you please tell me if the antiderivative of e^2x is e^2x/2 and if 2e^2x is e^2x/2? I got both of these answers by doing the opposite of taking the derivative. For example, the derivative of e^2x is e^2x(2), so wouldn't the antiderivative be e^2x/2?

    Thank you very much
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Could you please tell me if the antiderivative of e^2x is e^2x/2 and if 2e^2x is e^2x/2? I got both of these answers by doing the opposite of taking the derivative. For example, the derivative of e^2x is e^2x(2), so wouldn't the antiderivative be e^2x/2?

    Thank you very much
    how can the antiderivative of \frac 12e^{2x} be the same as that of 2e^{2x}? won't they differ by a factor of 4?


    \int e^{2x}~dx = \frac 12 e^{2x} + C

    we can get this, as you say, by reversing the differentiation operation. or the more standard way to do it is by a substitution of u = 2x. in addition, it is simple enough (if memorizing something is simple for me, it means it's simple, my memory is horrible) to memorize \int e^{kx}~dx = \frac 1k e^{kx} + C, where k is a constant, and k \ne 0

    now what do you think \int 2e^{2x}~dx is?
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    You can also memorize the rule that if F(x) +C= \int f(x) dx then \int f(ax+b) dx = \frac{1}{a}F(ax+B)+C if a\not = 0.
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    Quote Originally Posted by ThePerfectHacker View Post
    You can also memorize the rule that if F(x) +C= \int f(x) dx then \int f(ax+b) dx = \frac{1}{a}F(ax+B)+C if a\not = 0.
    You must be forgetting about linear functions. Why would you divide by the constant a in that case?
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    What are you taking about?
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    Quote Originally Posted by colby2152 View Post
    You must be forgetting about linear functions. Why would you divide by the constant a in that case?
    he is talking about linear functions. his conclusion follows by integration by substitution. that's how i do it as well. whenever i have a composite function in which the inner function is linear, i don't go through the trouble of integrating by substitution, i just know the integral will be the integral of the whole function divided by the derivative of the inner function--pretty much the direct opposite of the chain rule for derivatives (which is what the method of substitution models in general)
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    Quote Originally Posted by ThePerfectHacker View Post
    What are you taking about?
    Your rule fails if you let f(x) = x.
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    Quote Originally Posted by colby2152 View Post
    Your rule fails if you let f(x) = x.
    How?

    \int x dx = \frac{1}{1}F(1x+0)+C where F(x) = \int xdx = \frac{1}{2}x^2.
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    Quote Originally Posted by ThePerfectHacker View Post
    How?

    \int x dx = \frac{1}{1}F(1x+0)+C where F(x) = \int xdx = \frac{1}{2}x^2.
    Let a = 2, so F(2x) = 2x^2... so unless I am mistaken your rule would give an answer of just x^2
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    Quote Originally Posted by colby2152 View Post
    Let a = 2, so F(2x) = 2x^2... so unless I am mistaken your rule would give an answer of just x^2
    You are mistaken.
    \int (2x) dx = \frac{1}{2}F(2x)+C where F(x) = \int x dx = \frac{1}{2}x^2.
    Thus,
    \frac{1}{2}F(2x) = \frac{1}{2}\cdot \frac{1}{2}(2x)^2 = x^2
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