Question

I'm in need of assistance for getting e^2 as my answer to this question. I think I got the correct integral, with the parameters being from e to e^2 for the equation ln(x) - 1. However, I am unsure how to integrate this. Do I set u = ln(x) or u = ln(x) - 1?

Answer 1

@justjm can you provide assistance

Answer 2

is it the area of A or area of B

Answer 3

So yes you are right if the region would be A. You would need to find
\(\int_{e}^{e^{2}}\ln x-1~dx\)

You can distribute the integral:
\(\int_{e}^{e^{2}}\ln xdx-\int_{e}^{e^{2}}1dx\)

I'm sure you can do the second term, but for the first term with ln x, you need to use integration by parts and then use FTC II as you would

integration by parts formula: \(uv-\int_{ }^{ }v'du\)
let \(u=\ln x\)
\(du=\frac{1}{x}dx\)
let \(dv=dx\)
\(\int_{ }^{ }dv=\int_{ }^{ }dx\) so v=x
\(x\ln x-\int_{ }^{ }x\cdot\frac{1}{x}dx\)
\(x\ln x-x+C\)

Thats your antiderivative for \(\int_{ }^{ }\ln xdx\), then you'd continue as you would with FTC II. Remember to include the second term from the beginning, and also to neglect the +C when you do FTC II

Answer 4

Thanks, I got that integral in the end, but for whatever reason, when I plugged it into the calc, I kept getting e as my final answer, but then it took me a while to realize that the calculator was actually returning the wrong answer.

Answer 5

I plugged in the equation to the integral with lower bound being e and upper bound being e^2 and compared it to when I did it by hand, and I got different answers essentially.

Answer 6

Wait but isn't e the answer though

Answer 7

It's e^2, plug in e^2 and e into xln(x) - x
You will actually notice that if you use the integral function, the calculator returns just e

Answer 8

I found that by solving it by hand, plugging in e^2 returns e^2, and plugging in e returns 0. It's such a weird thing to see.

Answer 9

Hold on let me try by hand

Answer 10

woah I still get e by hand

Answer 11

Wait, now that I think about it, won't the integral be from e to e^2 with the equation xln(x) - 2x because of the x that was initially in the equation?

Answer 12

Is that for the antiderivative or integrand?

Answer 13

So like, we found the integral of ln(x) to be xln(x) - x, but then there's also the -1 that was never accounted for, where the antiderivative would be -x. So would it not be -2x?

Answer 14

I don't believe we can combine the two x's because we distributed the integral and then we are applying FTC II to the separate ones. Hold on, I will take a picture of my work and show you. @mhchen any thoughts

Answer 15

justjm's calculations look right to me

Answer 16

@justjm thanks so much bro, I was never aware of that. That clears up so much confusion I had.

Answer 17

Hold on man I'm not sure if it's a thing, it was never clarified explicitly but I've just seen that when definite integrals get distributed, textbooks plug in the limits separately