# Thread: having 2pi as the upper limit of an interval

1. ## having 2pi as the upper limit of an interval

i cant make this work
e.g.

$\int_0^2\pi \theta d\theta$

the pi is mean to be the upper limit of the integral, not inside the integrand
any helps?

also i cant find out how to do the inequality sign

2. Originally Posted by walleye
i cant make this work
e.g.

$\int_0^2\pi \theta d\theta$

the pi is mean to be the upper limit of the integral, not inside the integrand
any helps?

also i cant find out how to do the inequality sign
Use braces ^{...}: $$\int_0^{2\pi} \theta d\theta$$ gives $\int_0^{2\pi} \theta d\theta$

What inequality sign: <, > $\geq$, $\leq$ ....?

3. thankyou thankyou thankyou

hahaha "which inequality sign"
im sorry, its been a long day

i meant the .. = with a / thought it

4. $\int_0^{2\pi} \theta d\theta$

\int_0^{2\pi} \theta d\theta

5. $2 \not= 3$

ahh here we go, i found it

6. Originally Posted by pickslides
$\int_0^{2\pi} \theta d\theta$

\int_0^{2\pi} \theta d\theta
You should leave a small space in between the $\theta$ and the $d\theta$.

\int_0^{2\pi} \theta \, d\theta

gives

$\int_0^{2\pi} \theta \, d\theta$

7. new question sports stars

how do you do the multiplication sign?
haha sorry if these questions are annoyingly basic, but i figure you might be able to answer before i find it in a guide some where

8. $5 \times 3 \not= 14$

im learning

9. Originally Posted by walleye
new question sports stars

how do you do the multiplication sign?
haha sorry if these questions are annoyingly basic, but i figure you might be able to answer before i find it in a guide some where

10. Originally Posted by Prove It
You should leave a small space in between the $\theta$ and the $d\theta$.

\int_0^{2\pi} \theta \, d\theta

gives

$\int_0^{2\pi} \theta \, d\theta$
Good point, I like to write it like this

\int_0^{2\pi} \theta~d\theta

$\int_0^{2\pi} \theta~d\theta$

11. Originally Posted by walleye
$5 \times 3 \not= 14$

im learning

Also you can consider \neq

$
\neq
$