Question

http://img705.imageshack.us/img705/7006/captureuiu.png

Answer 1

Bloackd!

Answer 2

what?

Answer 3

Blocked*

Answer 4

You can't see the image?

Answer 5

Nah

Answer 6

ok
\[g(0)=0\]

Answer 7

because for one thing
\[\int_a^a f(t)dt=0\] always

Answer 8

The first answer was 100% correct, satellite. :) You have the others as well?

Answer 9

and here you have
\[g(x)=\int _0^xf(t)dt\] and thus
\[g(0)=\int_0^0f(t)dt=0\]

Answer 10

yeah sure

Answer 11

you are computing areas using the formula for triangles or rectangles

Answer 12

\[g(1)=\int_0^1f(t)dt\] and that area is 2 since you have a rectangle with base 1 and height 2

Answer 13

\[g(2)=\int_0^2f(t)dt\] and that area is 4+1=5. there is a square with base 2 and height 2, plus that triangle on top with base 2 and height 1

Answer 14

\[g(3)=\int_0^3f(t)dt=5+2=7\]

Answer 15

the 5 from the previous area plus that triangle with base 1 and height 4

Answer 16

\[g(4)=\int_0^4f(t)dt = 6.5\] and i know this was not asked for but it is the area above the curve which we know is 7 minus the area below which is .5

Answer 17

\[g(5)=\int_0^5f(t)dt=7-2=5\] again the area above minus the area below. i know this wasn't asked for either but i hope it is clear where i am getting these numbers from

Answer 18

Satellite, you poor guy...this is the most brilliant explanation I've seen on OpenStudy yet...the answers alone would suffice, mate! :) is there any way i can give you more than one medal?

Answer 19

and
\[g(6)=\int_0^6f(t)dt = 7-4-3\] again area above curve minus area below

Answer 20

no i appreciate what i assume is a compliment, and more to the point i hope it is entirely clear where these numbers are coming from. i am just computing the area below the graph and above the x - axis. if the graph is below the x - axis that area comes with a minus sign. good luck on the rest.

Answer 21

Wait, what about the interval/maximum value??

Answer 22

and yes it was definitely a compliment lol

Answer 23

you were on a roll there, i didnt mean to stop you !

Answer 24

well it is increasing where f is above the x - axis because there the areas are positive and so you keep adding

Answer 25

just as i did when computing them. once the area is below it is obviously decreasing as we saw. for example
\[g(6)<g(5)<g(4)\]

Answer 26

and finally the maximum value for
\[g(x)=\int_0^xf(t)dt\] would be at 3, because that is where f goes from being positive to negative, so you have reached the greatest area there. the rest will come in negative

Answer 27

You are brilliant, sir. I salute you