Question

Solve the equation.

Answer 1

\[\int\limits_{\ln2}^{x} \frac{ dx }{ \sqrt{e^x-1} } = \frac{ \pi }{6 }\]

Answer 2

interesting

Answer 3

How are you integrating with respect to the upper bound in your definite integral

Answer 4

The upper limit is x. We get a value of x from that equation.

Answer 5

do you know how to get \[\int \frac{\rm dx}{\sqrt{e^x - 1}}\]
??

Answer 6

just the indefinite integral

Answer 7

Working on that.

Answer 8

Yeah, substitue \[e^x-1 = t^2\]
Then,
\[dx = \frac{ 2t }{ 1+t^2 }dt\]

I hope now its easy.

Answer 9

Are you implying that the integral is constant with respect to changes in its upper bound

Answer 10

seems unlikely

Answer 11

@Hyun11 , we get a function in x after integrating.

Answer 12

to solve x...the indefinite integral must be taken first...since you can just put in x later on. it's for simplification

Answer 13

The upper bound of your integral is x!

Answer 14

yes.

Answer 15

Your free and bounded variables should not be the same, this is just bad ambiguous notation, change the variable your integrating with respect to, to $t$ or something.

Answer 16

assuming you have \[\int_2^x 2xdx\]
you can just solve int 2x first which is x^2 + c

then plug in x... (x^2)- (4)

something like that

Answer 17

would you write \[\sum_{k=1}^k k\]

Answer 18

Yeah, but in many books this is followed.
My integration is correct, right?

Answer 19

\[\int\limits_{a}^x f(x)dx\]

Answer 20

wonder how summations got in....anyway...let me see

Answer 21

is the same pellet

Answer 22

summation is used for discrete numbers...integration is for continuous functions...just because they both add means they're the same

anyway...the sub seems right

Answer 23

I am illustrating the ambiguity of his notation, not discussing partial sums.

Answer 24

Ok,thanks for help. I will search some books about this notation.

Answer 25

in improper integrals you denote it as \[\lim_{t \rightarrow \infty} \int_{-t}^{t} f(x)dx\]
so i don't see why having variable bounds is ambiguous

Answer 26

xlegendx, I don't know what you are talking about

Answer 27

having variable bounds isn't ambigious

Answer 28

then what are you implying to be ambiguous?

Answer 29

having the same variable bound, as the variable your integrating with is ambigious

Answer 30

\[\int\limits_{a}^x f(x) dx\]

Answer 31

would be tantamount to writing\[\sum_{k=1}^k k\]

Answer 32

it would be ambiguous if you think you're going to get a constant answer.

Answer 33

having an upper bound the same as your variable of integration doesn't make sense if there's no equal sign

Answer 34

I don't think you understand what I am saying, we seem to be talking about totally different things

Answer 35

i think we are