#1**+1 **

If you differentiate g(x) with respect to x, what do you get?

Please answer, it is a simple question.

Melody Jan 30, 2021

#5**0 **

I understand its finding the area between 2 and 3, and that I know that the points are (2,4), and (3,5), so would the area be 4.5? then -1 so 3.5?

SpaceTsunaml
Jan 30, 2021

#9**0 **

Keep going sorry, I already thought of this but my sleep deprived mind has been doing too many of the "Find the volume of a solid by rotating said 2d function around an axis too cope with a simple 3 point problem apparently.

SpaceTsunaml
Jan 30, 2021

#11**+1 **

\(\displaystyle \int_2^3(g'(x)-1)dx\\~\\ =\displaystyle \int_2^3(g'(x))dx+\displaystyle \int_2^3(-1)dx\\~\\ =\displaystyle \left[ g(x) \right]_2^3+\displaystyle \left[ -x \right]_2^3\)

you should be able to finish it from there. Tell me what you get.

Melody Jan 30, 2021

#14**+1 **

ahhhh, okay well my thought process was integral means the g(3) - g(2) + (-3+2). Is this wrong?

oh okayyy you had me freaked out for a sec LOL

SpaceTsunaml
Jan 30, 2021