6970
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1
2011-02-01T00:00:01
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<div>
<br />
I asked my integrating application to work out this integral:<br />
<br />
Integrate[1000000/(10 + (1000000 - 10)/2^(4*x)), x]<br />
<br />
I was given the answer<br />
<br />
(25000*Log[-32*(99999 + 16^x)])/Log[2]<br />
<br />
Try to evaluate this expression at $x=1$. What has gone wrong? Can
you work out the actual real answer and verify by
differentiation?<br />
<br />
</div>
<div class="framework">
<span style="font-style: italic;">Did you
know ... ?</span>
<br />
<br />
Technology is undoubtedly a very useful mathematical tool, but
needs to be used carefully and with some skepticism. If you use a
calculating aid it is very important to check that the answer
makes sense. Algebraic tools tend not to make errors, but often
introduce unnecessary complexity into an answer.</div>
<br />
</mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
The integrator (with proper formatting) gave the answer for my
integral $I$ as<br></br>
<br></br>
$$I = \frac{25000\log[-32(99999 + 16^x)]}{\log 2}$$<br></br>
<br></br>
The problem is that the logarithm of a negative number is complex.
However, the rules of logarithms still apply, so we see that<br></br>
$$<br></br>
I = \frac{25000}{\log 2}\Big(\log(-32) + \log(99999+16^x\Big)<br></br>
$$<br></br>
The $\log(-32)$ part, although complex, is a constant of
integration which has been chosen by the integrator. Thus, the
integral is of the form<br></br>
$$<br></br>
I = \frac{25000\log(99999+16^x)}{\log 2} + c<br></br>
$$<br></br>
This is perfectly real for real choices of $c$ and direct
computation shows that this differentiates down to our starting
function.<br></br>
<br></br>
<br></br></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
The integrator (with proper formatting) gave the answer for my
integral $I$ as<br></br>
<br></br>
$$I = \frac{25000\log[-32(99999 + 16^x)]}{\log 2}$$<br></br>
<br></br>
The problem is that the logarithm of a negative number is complex.
However, the rules of logarithms still apply, so we see that<br></br>
$$<br></br>
I = \frac{25000}{\log 2}\Big(\log(-32) + \log(99999+16^x\Big)<br></br>
$$<br></br>
The $\log(-32)$ part, although complex, is a constant of
integration which has been chosen by the integrator. Thus, the
integral is of the form<br></br>
$$<br></br>
I = \frac{25000\log(99999+16^x)}{\log 2} + c<br></br>
$$<br></br>
This is perfectly real for real choices of $c$ and direct
computation shows that this differentiates down to our starting
function.<br></br>
<br></br>
<br></br></mdoxml>
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Weekly Challenge 24: Suspicious Integrator
A weekly challenge: these are shorter problems aimed at Post-16
students or enthusiastic younger students. What has happened with
my online integrator?
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