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anonymous
 5 years ago
Consider the area between the graphs x+3y=1 and x+9=y2. This area can be computed in two different ways using integrals
First of all it can be computed as a sum of two integrals; Alternatively this area can be computed as a single integral
anonymous
 5 years ago
Consider the area between the graphs x+3y=1 and x+9=y2. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals; Alternatively this area can be computed as a single integral

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and the question for this statement?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we can integrate them both at the same time, or seperately and add the sums...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And I need a and b, as well as c

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the common line between them is the sum of there parts./2 right? gets an averge?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I also need f(x) and g(x) separately

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)=x/3 +1/3; g(x) = sqrt(x+9)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0[S] f(x) dx  [S] g(x) dx [S] f(x)  g(x) dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The fist thing you have to do is to find the intersection of the two lines. Secondly, you have to find the intersection of the two lines with the x axis. As the third step, you have to integrate the frist integral from the xcoordinate of the first line until the xcordinate of the intersection point. You have to do the same with the seconde line

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it doesnt provide an interval, and so assuming that it just wants the intersections is presumptuous; prolly true, but not accurate

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it could want the interval [0,root to the right]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im not sure how to do that. Sorry

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you get something that looks like this

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0without a stated interval, you are prolly correct in assuming the left and right intercepts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh ok. give me one second to thing about it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its got issues of being on the other side of the y axis; and youd have to split it into intervals from the common point to the left and right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0maybe +9 to get it all on the right side of the y axis?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and my line parts to steep , but its just an artist interpretation lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anyway, you have to split the total area in two sides.Carry out integration from intersection of the parable and the xaxis until the intersection of the parable and the line. They you have to integrate from the intersection point to the intersection of the line with the xaxis. That is how I would do that my friend

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got x=1 for c, but it keeps saying that it's wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the end point on the right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont know my friend. I got a little bit confused. Sorry :(
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