Question

http://www.webassign.net/mapleplots/6/2/c13f7764b81080be24cc7fa9af7565.gif

Answer 1

i have to elvaulate..g(0), g(3), g(6), g(9), g(12), g(15), and g(18)

Answer 2

What is g? The graph shows f :P

Answer 3

yes..but the assignment tells me to find g(x)...in other
Let g(x) =\[\int\limits_{0}^{x}\] f(t)dt

Answer 4

g(x) is the integral of f(t), so it is the total area under the curve between 0 and x. If the curve is negative at some point, then that counts as negative area. I would go ahead and evaluate it geometrically.

Answer 5

Have you tried finding the equation of the lines? For example the equation of the graph on the domain [0, 6) is f(t) = -x +3 I believe.

Answer 6

Sorry, f(t) = -t +3

Answer 7

Thus \[g(x) = \int\limits_{0}^{x}-t+3 dt\]
\[=-t ^{2}/2+3t ] from 0 \to x\] (sorry there's no way in the equation editor to type that properly)
\[=-x ^{2}/2+3x\]
so I'm thinking that you should be able to plug in values like g(0) for those x values in the domain of [0, 6).

Answer 8

Sorry, Throgg, but that's really overcomplicating things. Yes, it works, but it's not the simplest way to do it.

If you recognize that the integral corresponds to the area under the curve, then you can just use simple geometry to find what the area under the curve is.

For g(0), there's no area.
for g(3), the area is a triangle of base 3 and height 3 for an area of 1/2 * 3 * 3 = 4.5
for g(6), there is an additional triangle that is congruent, but it is under the x axis, so it is negative. The negative area is the same as the positive area, so the total area is 0.
For g(9), there is a second congruent but negative triangle formed, so there is -4.5 area.
for g(12), there are now four triangles formed of base 3 and height 3, but 2 of them are negative and 2 are positive, so the area is again 0.
for g(15), we can see that everything before x=12 cancels out to 0, so only the area from x =12 to x = 15 needs to be accounted for. This area can be broken down into a triangle of base 3 and height 3 along with a 3 by 3 rectangle for a total area of (1/2)*3*3 + 3*3 = 13.5
for g(18), we add to this two more 3*3 rectangles and another triangle of base 3 and height 3. So the total area is 13.5 + 3*3*2 + (1/2)*3*3 = 36

Kablam, sucka.