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

Question

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

amistre64 · Answer

and the question for this statement?

amistre64 · Answer

we can integrate them both at the same time, or seperately and add the sums...

anonymous · Answer

I need both please!

anonymous · Answer

And I need a and b, as well as c

amistre64 · Answer

the common line between them is the sum of there parts./2 right?  gets an averge?

anonymous · Answer

I also need f(x) and g(x) separately

amistre64 · Answer

f(x)=-x/3 +1/3; g(x) = sqrt(x+9)

amistre64 · Answer

[S] f(x) dx - [S] g(x) dx

[S] f(x) - g(x) dx

anonymous · Answer

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

amistre64 · Answer

it doesnt provide an interval, and so assuming that it just wants the intersections is presumptuous; prolly true, but not accurate

amistre64 · Answer

it could want the interval [0,root to the right]

anonymous · Answer

Im not sure how to do that. Sorry

amistre64 · Answer

you get something that looks like this

amistre64 · Answer

without a stated interval, you are prolly correct in assuming the left and right intercepts

anonymous · Answer

ahhh ok. give me one second to thing about it

amistre64 · Answer

its 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 · Answer

maybe +9 to get it all on the right side of the y axis?

amistre64 · Answer

and my line parts to steep , but its just an artist interpretation lol

amistre64 · Answer

kinda like this eh

anonymous · Answer

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

anonymous · Answer

I got x=1 for c, but it keeps saying that it's wrong.

anonymous · Answer

which c?

anonymous · Answer

the end point on the right

anonymous · Answer

I dont know my friend. I got a little bit confused. Sorry :(