for 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.
weloo
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.
1 answer