Brain Teaser - Area enclosed by a parabola and a line

[CR Haeger]

(09-13-2015 12:17 AM)Gerson W. Barbosa Wrote:  

(09-11-2015 04:57 PM)CR Haeger Wrote:  I tried to work the following problem fully by hand and then with CAS help from the HP Prime. Eventually I got to "solutions" on the Prime but it took a lot of variables creation and substitutions to get there. I am sure most here will make quick work of this.

The problem:
* Find the area enclosed by a function f(x)=x^2 and a line g(x) that intersects and is normal to f(x) for all x>0. Solve for area at x=+1.0.

* Write the function h(x) in terms of x only which describes the area enclosed by f(x) and g(x).

* Find the minimum value of h(x)

* h(x) = (4x² + 1)³/(48x³)

* 4/3 at x = 1/2

I've fount the normal lines at x = 1, x = 2 and x = 3, then generalized for X. (Picture later if important).
I've used a quadratic solver to generalize the limits of integration.
Too lazy to do the remaining by hand:

http://www.wolframalpha.com/input/?i=int...%29%29%2CX

http://www.wolframalpha.com/input/?i=sol...%29%3D%3D0

Regards,

Gerson.

Good work!

I struggled (by hand) determining the general expression for the lower limit of integration. Tried solving x^2-m*x-b = 0 for x where m and b were slope, intercept of line g(x). Your expression seems better simplified.

Also, the HP Prime placed some absolute value limits on my h(x) equation... Ill post what I calculated soon.