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
Stacey Warren - Expert brainly.com
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and the question for this statement?
we can integrate them both at the same time, or seperately and add the sums...
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And I need a and b, as well as c
the common line between them is the sum of there parts./2 right? gets an averge?
I also need f(x) and g(x) separately
f(x)=-x/3 +1/3; g(x) = sqrt(x+9)
[S] f(x) dx - [S] g(x) dx
[S] f(x) - g(x) dx
The 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 x-coordinate of the first line until the x-cordinate of the intersection point. You have to do the same with the seconde line
it doesnt provide an interval, and so assuming that it just wants the intersections is presumptuous; prolly true, but not accurate
Anyway, you have to split the total area in two sides.Carry out integration from intersection of the parable and the x-axis 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 x-axis. That is how I would do that my friend
I got x=1 for c, but it keeps saying that it's wrong.
the end point on the right
I dont know my friend. I got a little bit confused. Sorry :(