What is the sign of the integral for the function y=log_{10}x from x=2 to x=4?

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SUMMARY

The integral of the function y=log10x from x=2 to x=4 is calculated using the change of base formula and integration by parts. The correct evaluation leads to the expression -1/(4ln10), approximately equal to -0.1086. Despite the negative result, the function remains positive in the specified range, indicating a misunderstanding in the integration process rather than an error in the calculation itself. The integral's negative value arises from the method of integration applied, specifically the integration by parts technique.

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Homework Statement


Find
\int_2^4{log_{10}xdx

Homework Equations


Use the change of base formula
log_ba=\frac{log_ca}{log_cb}

The Attempt at a Solution


Using the formula:
\frac{1}{ln10}\int_2^4lnxdx

=\frac{1}{ln10}[\frac{1}{x}]_2^4

=\frac{1}{ln10}(\frac{1}{4}-\frac{1}{2})

=\frac{-1}{4ln10} \approx -0.1086

What my problem is, is that the integral is giving a negative value, but the function of y=log_{10}x is positive for 2\leq x\leq 4.

So is it that I've made a mistake in calculating this integral, or is the negative value I'm getting legitimate? At this point I would need to reconsider when the integral is ever negative.
 
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\int ln(x)dx = xln(x) - x
 
Mentallic said:

Homework Statement


Find
\int_2^4{log_{10}xdx

Homework Equations


Use the change of base formula
log_ba=\frac{log_ca}{log_cb}

The Attempt at a Solution


Using the formula:
\frac{1}{ln10}\int_2^4lnxdx

=\frac{1}{ln10}[\frac{1}{x}]_2^4

What've you done here? You seem to have differentiated instead of integrated. As Random Variable mentions above, the 'trick' to integrating the logarithm function is to write it as 1.ln(x) and use parts.
 
Oh hehe, oops xD

Uhh... I still don't understand how you obtained that result Random Variable. Can you guys please explain this
cristo said:
the 'trick' to integrating the logarithm function is to write it as 1.ln(x) and use parts.
a little more?
 
Mentallic said:
Can you guys please explain this a little more?

You can write ln(x) as 1.ln(x), to which you can then apply integration by parts. Let u=ln(x), dv=1dx, which will give du=1/x dx, v=x. Then, recall that

\int u dv=uv-\int v du\Rightarrow\int \ln x dx=x\ln x-\int x\frac{1}{x} dx = x\ln x - x (+C)\,.
 
This has never been done in class before. Maybe we were meant to experiment or just know this result by natural instinct? :smile: Thanks.
 
Do you mean that this integral hasn't been done in class, or the technique of integration by parts hasn't been done in class?
 
Integration by parts. And this integral hasn't been done either or else it wouldn't make sense for the teacher to throw it in an assignment soon after showing us how to solve it (which would mean she would probably have taught us integration by parts at that point :-p).
 

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