Definite intergral with one variable trying to solve for constant

The problem is i have this definite integral which i need to integrate and then solve for P:

F = P*x^2-5000*x^3/((x/26 + 1/10)^3/300 - (x/26 + 23/250)^3/375)

between the limits of 0 and 2.6 but i am unsure on how to get it to solve for P.

syms x P

F = P*x^2-5000*x^3/((x/26 + 1/10)^3/300 - (x/26 + 23/250)^3/375);

Fint = int(F, x, 0, 2.6)

So, integrate the function between 0 and 2.6 with x being the varible to get an answer for P?

Any help will be much apprieciated

Answer 1

[Star Strider]

There is no closed-form, analytic expression fot the integral. In order to solve for the value of ‘P’, it will be necessary to know the value of the integral.

This assumes that the value of the integral is 0:

syms x P 
F = P*x^2-5000*x^3/((x/26 + 1/10)^3/300 - (x/26 + 23/250)^3/375);
Fint = int(F, x, 0, 2.6);
P = solve(Fint,P)                                                   % Assumes Value Of The Integral Is 0

producing:

P =
1742302746.6832036018225596059422

.

[John D'Errico]

Actually, there is. It won't be anything pretty, because it probably does something like a partial fraction expansion of the second term, and two of the roots of the denominator polynomial are complex.

vpa(solve((x/26 + 1/10)^3/300 - (x/26 + 23/250)^3/375))
 
ans =
 
- 2.5011523457776391810634189976684 + 0.059933847814197846979738111724203i
- 2.5011523457776391810634189976684 - 0.059933847814197846979738111724203i
                                        -5.2936953084447216378731620046633
 

So the result is a lot of ugly looking crap.

pretty(vpa(int(F,x),5))
log(6.7785e+23 x + 3.5883e+24) 2.5064e+12 - 1.3182e+11 x + log(x (5.7481e+26 - 3.2008e+26i) + 1.4185e+27

   - 8.3503e+26i) (- 5.7461e+11 - 6.1387e+12i) + log(x (5.7481e+26 + 3.2008e+26i) + 1.4185e+27 + 8.3503e+26i) (- 5.7461e+11

                               3
   + 6.1387e+12i) + 0.33333 P x

But it has a closed form expression.

[Tanay Manoj Vaghadia]

Thanks for the help but the value found seems awfully big as i was expecting the around 8000 Newtons i would assume its something with the equation?

[Star Strider]

Our pleasure!

John — Thank you!

‘... i would assume its something with the equation?’

Quite possibly.

Answer 2

[Steven Lord]

Let's look at what your function looks like on the interval [0, 2.6] when P is 8000.

F = @(x, P) P*x.^2-5000*x.^3./((x/26 + 1/10).^3/300 - (x/26 + 23/250).^3/375);
fplot(@(x) F(x, 8000), [0, 2.6])