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Why is the area under this curve negative?

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Find your uni forum to get talking to other applicants, existing students and your future course-mates 27-07-2015
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    y=\frac{1}{2}^x

    When I use a graph sketcher to sketch it, there is no part that goes under the x-axis.

    \int^\infty_0 \frac{1}{2}^x\ dx = -1.44

    I know that it's negative because \int \frac{1}{2}^x\ dx = \frac{\frac{1}{2}^x}{\ln\frac{1}  {2} } + C

    and ln(1/2) is negative, so my answer will be negative. But I don't understand why there is a negative number there. Is it just something I have to ignore? or is there a reason for it?

    Wolfram alpha calculates it as postive so maybe my working is wrong? If I'm being really stupid, I'm really sorry
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    Surely the numerator tends to -1
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    (Original post by hello calum)
    y=\frac{1}{2}^x

    When I use a graph sketcher to sketch it, there is no part that goes under the x-axis.

    \int^\infty_0 \frac{1}{2}^x\ dx = -1.44

    I know that it's negative because \int \frac{1}{2}^x\ dx = \frac{\frac{1}{2}^x}{\ln\frac{1}  {2} } + C

    and ln(1/2) is negative, so my answer will be negative. But I don't understand why there is a negative number there. Is it just something I have to ignore? or is there a reason for it?

    Wolfram alpha calculates it as postive so maybe my working is wrong? If I'm being really stupid, I'm really sorry
    \int^\infty_0 \frac{1}{2}^x\ dx = \int^\infty_0 e^{x\ln 0.5} dx = [\frac{e^{xln0.5}}{\ln 0.5}]^{\infty}_{0} = 0 - \frac{1}{\ln 0.5}= -(\ln 0.5)^{-1} > 0

    unless I've made an error. So I think it's your working.
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    its -1/Log[1/2] = 1.4426
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    Substitute in Remember that there's another minus sign as the 0 is the lower limit (and the infinite limit disappears).
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    omg. I was putting them in the wrong way round :/ Sometimes I just do stuff the wrong way round for some reason
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    (Original post by atsruser)
    \int^\infty_0 \frac{1}{2}^x\ dx = \int^\infty_0 e^{x\ln 0.5} dx = [\frac{e^{xln0.5}}{\ln 0.5}]^{\infty}_{0} = 0 - \frac{1}{\ln 0.5}= -(\ln 0.5)^{-1} > 0

    unless I've made an error. So I think it's your working.
    How does the evaluation at infinity equal 0?
    x \ln{0.5} = \ln{0.5^x} which tends to \ln0 as x tends to infinity, right?

    I get that e^{\ln X} = X but surely e^{\ln0} = 0 isn't valid, as what on earth is \ln 0?

    EDIT: Nevermind, I was being stupid. Of course, \ln0.5 is negative so e^{\infty \ln0.5} = e^{-\infty \ln2} = 0. Derp. :facepalm:

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