calculus: evaluate the integral, from 5 to e^1, of (1/x) dx

[venialove]

Evaluate \(\displaystyle \int_{5}^{e^{1}}\frac{1}{x}dx\)

 

[galactus]

The antiderivative of 1/x is ln(x). This is very straight forward.

 

[venialove]

Re: calculus help

It is actually posted like this:
\[ \int_1^5e 1(x)\,dx.\]

This is my work:
Doesn't 5e becomes 1/5e because I can move it in front of the integral?

 

[galactus]

It is actually posted like this:
\[ \int_1^5e 1(x)\,dx.\]

That's my point, that is incorrect. Use 'preview' to check your display before hitting 'submit'.

This is my work:
Doesn't 5e becomes 1/5e because I can move it in front of the integral?

\(\displaystyle \int_{1}^{5e}\frac{1}{x}dx\) or \(\displaystyle \int_{1}^{5e}xdx\)

Is that correct?.

Please post the correct set up. I am not going to attempt a guess.

 

[venialove]

Evaluate the integral, from 1 to 5e, of (1/x) dx

This is the set up \(\displaystyle \int_{1}^{5e}\frac{1}{x}dx\)

 

[galactus]

Re: Evaluate the integral, from 1 to 5e, of (1/x) dx

One must know that the antiderivative of 1/x is ln(x), then it's all downhill.

Then we get \(\displaystyle \int\frac{1}{x}dx=ln(x)\)

Now, use the given limits of integration and use the first fundamental theorem of calculus.

\(\displaystyle ln(5e)-ln(1)\)

We can use a log identity on the first part because ln(1)=0.

Remember that ln(ab)=ln(a)+ln(b). So we get:

\(\displaystyle ln(5e)=ln(5)+ln(e)=ln(5)+1\)

And that is our solution.

See?.

 

[venialove]

Yes indeed! Thank you so much. You're Awesome!
God Bless