Calculus - area between 2 curves and perimeter enclosing the area?

let y1 = x^2 - 10x; and y2 = 4x - x^2;

Then y1' = 2x - 10; and y2' = 4 - 2x;

y1 is a parabola with vertex below x-axis opening upwards ( U shaped), and y2 is a parabola with

vertex above x-axis opening downwards (inverted). putting y1=y2 we get x=0,7 are the two

intersection points x=7 point is below x-axis. Inverted parabola y2 meets x axis at x=0,4 (got

by putting y2=0)

Thus required area A = Integrate 0 to 4 (y2 * dx) + | Integrate 0 to 4 (y1 * dx) | +

| Integrate 4 to 7 ( (y1 - y2) * dx ) |

= 32/3 + | -176/3 | + | -45 |

= 32/3 + 176/3 +45

= 343/3 square units (ANSWER)

since two parabolas opening upwards and downwards intersect, therefore total enclosed length is

symmetrical

Thus required length S =2 * ( Integrate 0 to 7 (sqrt( 1+ y2' ^2 ) * dx) )

= 2 * ( Integrate 0 to 7 (sqrt( 1 + y1' ^2 ) * dx) )

= 2 * 30.52102855

= 61.0420571 (there is decimal point after 61)

= 61.0420571 units (ANSWER)

First we find the x-coordinates of where the two graphs cross : for this, x^2 - 10x = 4x - x^2 or

2x^2 - 14x = 0, 2x(x - 7) = 0 so x = 0 or x = 7. The fact that the first graph is concave up (i.e. is U shaped) and the other is an inverted U means that the "upper" curve in the region in question is y = 4x - x^2, and the lower is y = x^2 - 10x - for convenience I'll refer to these as f(x) and g(x) respectively, so that the area enclosed = INT{f(x) - g(x)}dx from x = 0 to x = 7 = INT{14x - 2x^2)dx =

[7x^2 - (2/3)x^3] from x = 0 to x = 7, and since the expression = 0 at x = 0, the area is 7*7^2 - (2/3)*7^3 = 7^3(1 - 2/3) = 7^3/3 = 343/3 units^2. For the perimeter of the shape we will have to calculate it in two parts (i) the curve y = 4x - x^2 and (ii) the curve y = x^2 - 10x. In both cases, if s denotes the arc length then ds/dx = sqrt{1 + (dy/dx)^2}. For y = 4x - x^2, dy/dx = 4 - 2x so ds/dx = sqrt{1 + (4 - 2x)^2} = sqrt[4x^2 - 16x + 17] and this can be written sqrt{1 + 4(x - 2)^2}. I'm running out of time, so I'll have to leave that to you, but the other arc length is ds/dx = sqrt[ 1 + (2x - 10)^2] = sqrt {1 + 4(x - 5)^2} so once you find a suitable substitution for the first, it will be very similar for the second. Good luck with the arc lengths.

For issues like this coping with looking the biggest section, the biggest achievable section is often a sq.. hence a sq. with a edge of 100ft has aspects of a hundred/4 = 25ft, giving a community of 25² = 625 sq. ft. (the stuff relating to the wall does not remember)