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October 25, 2014

October 25, 2014

Posted by **weloo_volley** on Tuesday, May 8, 2012 at 6:11pm.

Consider the two parabolas :y1=2x^2-3x-1 and y2=x^2+7x+20.

(a) Find the points of intersection of the parabolas and decide which one is greater than the other between the intersection points.

(b) Compute the area enclosed by the two parabolas.

(c) Use Mathcad to draw the graphs of the parabolas and find their intersection points graphically .

(d) Solve the differential equation dy/dx = -exp(5-x)+3/5x^2 ; y(5)=3.

- math -
**Reiny**, Tuesday, May 8, 2012 at 6:57pmfor their points of intersection...

2x^2 - 3x - 1 = x^2 + 7x + 20

x^2 - 10x - 21 = 0

x^2 - 10x + 25 = 21 + 25 , I completed the square

(x-5)^2 = 46

x-5 = ± √46

x = 5 ± √46

so x = 5+√46 and x = 5-√46

or appr. 11.8 and -1.8

so I will take any value of x between these two values, x = 0 seems like a good choice, and find their y values

if x=0, y1 = -1, and if x=0 , y2 = 20

so the parabola y2 = .... is greater for all values of x between the two values we found

So the height of the region is

x^2+7x+20-(2x^2-3x-1) = -x^2 + 10x + 21

Area = ∫(-x^2 + 10x + 21) dx from x = 5-√46 to 5+√46

= [(-1/3)x^3 + 5x^2 + 21x] from x = 5-√46 to 5+√46

= ...

I will let you do the button-pushing arithmetic

Obviously we cannot demonstrate Mathcad on this webpage.

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