Posted by TOMO on .
Consider the area between the graphs x+4y=14 and x+7=y^2. This area can be
computed in two different ways using integrals.
First of all it can be computed as a sum of two integrals
They ask to use two integrals so i put f(x) from 7 to 2 which is correct
but for g(x) i put 2 to 14 for some reason 14 is wrong. I also put
f(x)=sqrt(x+7) and g(x)= (14x)/4 and both are wrong wrong. I got
everything else correct except for these and I don't what I did wrong.

MATH 2B Calculus 
jai,
first, graph the two equations (in one cartesian plane)
then get the points of intersection:
x+7=y^2 *this is the second equation
x=y^27
substitute this to the first:
y^27+4y=14
y^2+4y21=0
(y+7)(y3)=0
y=7 and y=3
substitute these back to obtain corresponding values of x:
*if y=7,
x=(7)^27=42
*if y=3,
x=(3)^27=2
therefore, points of int are (42,7) and (2,3)
looking at the graph, i suggest you do vertical strips (that is, dx),, divide the whole area into region 1 (left side) and region 2 (right side),, after you do this, get the boundaries.
Region 1:
for x: the boundaries are the graph of the parabola (that is x=y^27) and the xcoord of the first point of int (x=2)
for y: since parabola is symmetric with respect to xaxis, y is from 3 to 3.
the area of region 1 is:
integral[from 3 to 3](integral[from y^27 to 2] dx)dy)
*note: this is double integral since A=dxdy
Region 2:
for region 2, i suggest you do horizontal strips (that is, dy) ,,then get the boundaries:
for x: from xcoord of first point of int (x=2) to xccord of 2nd point of int (x=42)
for y: from the parabola (y=sqrt(x+7)) to the line (y=(14x)/4)
the area of region 2 is:
integral[from 2 to 42](integral[from sqrt(x+7) to (14x)/4] dy)dx)
i'll leave the integration calculation to you,, add the areas and you'll finally get the whole area.
so there,, sorry for long explanation,,
i hope i was able to help,, =)