Question

Find the area of the region under the curve over the prescribed interval

y=2/(1+t^2) on [0,1]

Answer 1

i keep geting 90, but my online homework think keeps telling me otherwise

Answer 2

\[t=\tan x\]is that what you did?

Answer 3

well i got that the integral was 2(tan^-1(x))

Answer 4

the 2/(1+t^2) is equates to arctan(t)
so 2 * arctan(t) from 0 to 1

Answer 5

which 2 times 45 is 90 and 0 is just 0

Answer 6

do you need the answer in degrees or radians?

Answer 7

im pretty sure it doesnt matter. what ever form you put it in, the system will see that its equivalent.

Answer 8

\[2(\tan^{-1}1-\tan^{-1}0)\]

Answer 9

can you take a screenshot of the question

Answer 10

right thats what i got

Answer 11

no, it definitely matters
and it's almost certainly radians

Answer 12

so what is the answer in radians?

Answer 13

it does not specify

Answer 14

who is "it" ?

Answer 15

@m.auld64 do you remember your special values in radians from trig?
that's what you can use to solve this problem exactly

Answer 16

take a screenshot

open the window and hit the PrtSc button in the top right-hand part of your keyboard

then open up paint and paste it in (shortcut ctrl + v)
then upload

Answer 17

lol my computer. but i just put my calculator into radians and i got 1.57 and it accepted that. you were right turing test

Answer 18

it \(always\) in radians in calculus unless it says otherwise.
if not, you have to do a u-sub for the integral and... it's uglier

Answer 19

yeah, they are the same thing but most everything wants the answer in radians, unless otherwise specified

it will usually say that it wants it in degrees, otherwise it's usually radians

Answer 20

they are the same answer, you can check by doing

1.57 * (pi / 180)

Answer 21

right, but you need to change the units to radians, so...

Answer 22

same thing

Answer 23

but turing is right, the integration will be in rad