Question

calculus improper integral...

Answer 1

\[\int\limits_{2}^{\infty} \frac{ 1 }{ 1+x^2 } dx\]

Answer 2

what is the question
the integration?
evaluating the limit part after the integration?

Answer 3

determine if the improper integral converges. If it does, evaluate it

Answer 4

same question
what part are you having problems on
the integration?
or
evaluating the limit part after the integration?

Answer 5

both

Answer 6

the antiderivative of 1/(x^2+1) should be well known to you

Answer 7

do you know the derivative of arctan(x) ?

Answer 8

oh the integral is arctan(x)

Answer 9

\[\int\limits \frac{1}{x^2+1} dx=\arctan(x)+C \\ \int\limits_2^\infty \frac{1}{x^2+1} dx=\lim_{z \rightarrow \infty} (\arctan(x)|_2^z)\]
right now plug in the upper and lower limit
and then evaluate the limit

Answer 10

\[\arctan(2)-\arctan(\infty)\]

Answer 11

well you should plug in upper limit then minus plug in lower limit

Answer 12

and you aren't actually suppose to plug in infinity because infinity isn't a number

Answer 13

\[\arctan(b)-\arctan(2)\]

Answer 14

\[\lim_{z \rightarrow \infty}(\arctan(z)-\arctan(2))\]

Answer 15

or use b

Answer 16

whatever
now evaluate the limit

Answer 17

how do i do the arctan(z) part?

Answer 18

have you ever evaluated limits before?

Answer 19

this should have came before integration actually :p

Answer 20

z goes to infinity so arctan(z) goes to....

Answer 21

infinity?

Answer 22

no...

Answer 23

do you know what the graph of f(x)=arctan(x) looks like?

Answer 24

yes

Answer 25

then you should see that arctan(x) aka the y values are approaching a certain number as x gets super large