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Why is the area under this curve negative?

y=12xy=\frac{1}{2}^x

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

∫0∞12x dx=−1.44\int^\infty_0 \frac{1}{2}^x\ dx = -1.44

I know that it's negative because ∫12x dx=12xln⁡12+C\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 :frown:
Reply 1
Avatar for TenOfThem
TenOfThem
18
Surely the numerator tends to -1
Reply 2
Avatar for atsruser
atsruser
11
Original post by hello calum
y=12xy=\frac{1}{2}^x

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

∫0∞12x dx=−1.44\int^\infty_0 \frac{1}{2}^x\ dx = -1.44

I know that it's negative because ∫12x dx=12xln⁡12+C\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 :frown:


∫0∞12x dx=∫0∞exln⁡0.5dx=[exln0.5ln⁡0.5]0∞=0−1ln⁡0.5=−(ln⁡0.5)−1>0\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.
its -1/Log[1/2] = 1.4426
Substitute in :smile: Remember that there's another minus sign as the 0 is the lower limit (and the infinite limit disappears).
Reply 5
Avatar for hello calum
hello calum
OP
10
omg. I was putting them in the wrong way round :/ Sometimes I just do stuff the wrong way round for some reason :frown:
Original post by atsruser
∫0∞12x dx=∫0∞exln⁡0.5dx=[exln0.5ln⁡0.5]0∞=0−1ln⁡0.5=−(ln⁡0.5)−1>0\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?
xln⁡0.5=ln⁡0.5xx \ln{0.5} = \ln{0.5^x} which tends to ln⁡0\ln0 as x tends to infinity, right?

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

EDIT: Nevermind, I was being stupid. Of course, ln⁡0.5\ln0.5 is negative so e∞ln⁡0.5=e−∞ln⁡2=0e^{\infty \ln0.5} = e^{-\infty \ln2} = 0. Derp. :facepalm:
(edited 11 years ago)

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