Question

@DanJS

Answer 1

from what you said before, it would be 4 right? im not sure

Answer 2

I mean ill do the first one first

Answer 3

Where should X=a be placed to divide the region into a half area...

Answer 4

\[\int\limits_{-1}^{0}\sqrt{x+1}dx = 2/3\]

Answer 5

That is the total area above the x axis and below the curve. right

Answer 6

Now we need to find the x vertical line, that will divide that area in half, namely area of 2/6

Answer 7

So we set up an integral from -1 to a, over the same region, and set that equal to half the area 2/6

Answer 8

\[\int\limits_{-1}^{a}\sqrt{x+1}dx = \frac{ 2 }{ 3 }(a-1)^{3/2}\]

Answer 9

we need that expression to equal half the area, 2/6

Answer 10

\[\frac{ 2 }{ 3 }(a+1)^{3/2} = \frac{ 2 }{ 6 }\]

Answer 11

sorry, the integral from -1 to a, is (a+1) , i wrote a minus

Answer 12

If you solve that for a, you get

a = -0.37

Answer 13

A vertical line at x = -0.37 will divide the regions area in half

Answer 14

Here it is

Answer 15

Does it make sense?

Answer 16

i got lost for a second, im reading it again

Answer 17

k
First Find the area under the curve from -1 to 0, that is the total area. It is 2/3

Half of that area is 2/6

Need to integrate from -1 to some line at x=a, so that the area is 2/6

Answer 18

When you integrate from -1 to a, you get that expression F(a) - F(-1), with a in it.
2/3(a+1)^(3/2)

you then set that equal to half of the area 2/6 and solve for a

Answer 19

ohh yeah i understand it(:

Answer 20

cool

Answer 21

so now if you did the integral from -1 to that 'a', then from 'a' to 0 , they should be the same number, half the area

Answer 22

if you wanted to do a check

Answer 23

ohh, i didnt know i could do that to double check my answer

Answer 24

yeah, for that other prob, open a new thread , i have a graph ready

Answer 25

okay(: