Question

Calculus question

Answer 1

@geerky42 do i need do to the integral and then do F(1)?

Answer 2

\[\int\left(\dfrac{1}{x^2}-2\right)dx = \int \dfrac{1}{x^2}dx-\int2dx=\int x^{-2}dx-\int2dx\]

Answer 3

just take integral then find F(1) then solve for C

Answer 4

Do you know what to do from here?

Answer 5

@geerky42 so i got the integral which is
-2 x-1/x+C

Answer 6

set equal to 0, to solver for C?

Answer 7

no, see, you now have \(F(x)=2x-\dfrac{1}{x}+C\)
and you are given that F(1)=-1. so plug in 1 for x and set equal to -1 then solve for C.

Answer 8

ohh ok

Answer 9

Im getting 2 as the answer :/

Answer 10

actually -2

Answer 11

Are you sure you plugged in and set equal equal to -1 correctly? you should have this: \(\color{red}{-1} =2\cdot\color{red}1-\dfrac{1}{\color{red}1}+C\)

Answer 12

ok so you have C=-2.

now plug it in C in \(F(x) = 2x- \dfrac{1}{x} + C\)

So you should have \(F(x) = \boxed{2x-\dfrac{1}{x}-2}\)

Answer 13

nope :(

Answer 14

-2x, not 2x

Answer 15

im still getting it wrong even changing 2x to -2x

Answer 16

wait a sec it is actually
-2x-(1/x)+ 2

Answer 17

i got it

Answer 18

@OOOPS thank you to you too

Answer 19

:)

Answer 20

right, i somehow thought 2 was positive in integration... sorry

so you have \(F(x)=-2x-\dfrac{1}{x}+C\)

plug in 1 for x and set it equal to -1. solve for C again. what do you get?

Answer 21

i got -2x-(1/x)+2

Answer 22

omg my internet sucks...

Answer 23

so does that make sense? everything is good?

Answer 24

yeah