Why is the area under this curve negative?
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Why is the area under this curve negative?
When I use a graph sketcher to sketch it, there is no part that goes under the x-axis.
I know that it's negative because
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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Re: Why is the area under this curve negative?(Original post by hello calum)
When I use a graph sketcher to sketch it, there is no part that goes under the x-axis.
I know that it's negative because
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](http://www.thestudentroom.co.uk/latexrender/pictures/9f/9f488b9854d6988af707afcadf66a122.png "\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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Re: Why is the area under this curve negative?How does the evaluation at infinity equal 0?
which tends to 
as x tends to infinity, right?
I get that
but surely 
isn't valid, as what on earth is 
?
EDIT: Nevermind, I was being stupid. Of course,
is negative so 
. Derp. 
Last edited by Astronomical; 04-08-2012 at 23:29.
Remember that there's another minus sign as the 0 is the lower limit (and the infinite limit disappears).
