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

[Mentallic]

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.

 

[Random Variable]

$$\int ln(x)dx = xln(x) - x$$

 

[cristo]

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.

 

[Mentallic]

Oh hehe, oops xD

Uhh... I still don't understand how you obtained that result Random Variable. Can you guys please explain this

the 'trick' to integrating the logarithm function is to write it as 1.ln(x) and use parts.

a little more?

 

[cristo]

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

$$\int u dv=uv-\int v du\Rightarrow\int \ln x dx=x\ln x-\int x\frac{1}{x} dx = x\ln x - x (+C)\,.$$

 

[Mentallic]

This has never been done in class before. Maybe we were meant to experiment or just know this result by natural instinct? Thanks.

 

[Mark44]

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?

 

[Mentallic]

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 :tongue:).