Area Under the curve where you're shown the curve

scrum

Junior Member
Joined
Oct 11, 2007
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55
There are three of these on my homework that i have partially solved, but i can't quite get the concept of them. I think I may have a concept wrong that is causing me to mess up.
This is the first one, I solved it without a problem
_2-prob10-prob5_2_34.gif

174487img1.gif
= 4 because area of the triangle
= -2pi because it's a negative half circle
174487img3.gif
= 4-2pi+.5 by the same concept
This is a similar one that asks different questions
174487img1.gif

graph of f.
_2-prob15-prob5_3_2.gif

g(0) = 0 , correct.
i got this by plugging y = -1/2x^2 to grapher on my mac, then looking at the graph next to it's derivative and making the derivative look like the line on that graph.

g(1) = 1/2 , correct, because y = -1/2x^2+1x is 1/2 at that point and the first part of the function looks like the derivative of that
g(2) = I don't know how to cope with the "corner" in the graph.
g(3) = -4.5, incorrect. using the same logic as 0 and 1 i got -4.5 using grapher but this is wrong
g(4) =
g(5) =
g(6) =
g(7) =
by the way we're meant to do this without using computers. I think it's conceptual.
 
In the second one; \(\displaystyle g(2)=-1\) because part of the graph is below the x-axis,

Really, just look at AREA.
 
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