MATH 2B Calculus

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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)= (14-x)/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 -

    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^2-7
    substitute this to the first:
    y^2-7+4y=14
    y^2+4y-21=0
    (y+7)(y-3)=0
    y=-7 and y=3
    substitute these back to obtain corresponding values of x:
    *if y=-7,
    x=(-7)^2-7=42
    *if y=3,
    x=(3)^2-7=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^2-7) and the x-coord of the first point of int (x=2)
    for y: since parabola is symmetric with respect to x-axis, y is from -3 to 3.

    the area of region 1 is:
    integral[from -3 to 3](integral[from y^2-7 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 x-coord of first point of int (x=2) to x-ccord of 2nd point of int (x=42)
    for y: from the parabola (y=sqrt(x+7)) to the line (y=(14-x)/4)

    the area of region 2 is:
    integral[from 2 to 42](integral[from sqrt(x+7) to (14-x)/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,, =)

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