Given O(0,0),P(2,3), and Q(5,9) lie on the curve, find `int_2^5ydx+int_3^9xdy` :

**I'm not at all sure that you can assume that the curve is a parabola.** (Through any three noncollinear points there is exactly one quadratic function -- but this might be an exponential, a higher degree polynomial, etc...)

`int_2^5ydx` is a the area under the curve and above the x-axis from x=2 to x=5, while `int_3^9xdy` is the area above the line y=3, below the line y=9, and bounded by the y-axis and the curve.

When you combine these areas you get overlapping rectangles as seen in your diagram. One rectangle is 6x5 (the horizontal rectangle) and the other is 3x9. The overlap has dimensions 3x6.

**The area is 30+27-18=39 sq units.** (You can cut the figure into two rectangles and add their areas if you do not like the idea of the overlap.)

Let y=f(x)=`ax^2+bx+c`

f(x) posses through (0,0),(2,3) and (5,9).Let us determine the constats a,b and c .

`0=axx0+bxx0+c`

`c=0`

`3=4a+2b`

`9=25a+5b`

solving these system of equatios ,we have

a=1/10 ad b=13/10

Thus `y=f(x)=(1/10)x^2+(13/10)x`

`10y+169/4=(x+13/2)^2`

`x=-(13/2)+sqrt(10y+169/4)` (In fig it is in first quadrant)

`int_2^5ydx+int_3^9xdy=int_2^5((x^2+13x)/10)dx+int_3^9(-(13/2)+sqrt(10y+169/4))dy`

`=1/10{x^3/3+13x^2/2}_2^5+{-13x/2}_3^9+int_3^9sqrt(10y+169/4)dy`

`=1/10{125/3+13xx25/2-8/3-13xx4/2}-13/2{9-3}+1/15{(10y+169/4)^(3/2)}_3^9`

`=39 units`

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