用积分求解方程

Question

我正在尝试在Python中求解一个方程，其中未知数被“圈定”在一个本身定义在另一个变量上的积分内。

我尝试了不同的编码方式，但到目前为止没有成功...

这是我尝试过的最简单的代码：

INSERT INTO TRIP VALUES( 1, 10001, 'PKR768', '2015-01-12' );
INSERT INTO TRIP VALUES( 2, 10001, 'SYF777', '2015-02-20' );
INSERT INTO TRIP VALUES( 3, 10001, 'KKK007', '2015-03-12' );
INSERT INTO TRIP VALUES( 4, 10001, 'PKR768', '2015-06-29' );
INSERT INTO TRIP VALUES( 5, 20002, 'PKR768', '2015-01-12' );
INSERT INTO TRIP VALUES( 6, 10002, 'SYF777', '2015-02-20' );
INSERT INTO TRIP VALUES( 7, 30005, 'KKK007', '2015-03-12' );
INSERT INTO TRIP VALUES( 8, 10001, 'PKR768', '2015-01-13' );
INSERT INTO TRIP VALUES( 9, 10002, 'QRT834', '2015-09-17' );
INSERT INTO TRIP VALUES(10, 30005, 'KKK007', '2015-12-15' );
INSERT INTO TRIP VALUES(11, 10003, 'SST005', '2016-01-23' );
INSERT INTO TRIP VALUES(12, 10002, 'PKR768', '2016-03-12' );
INSERT INTO TRIP VALUES(13, 20002, 'QRT834', '2015-04-23' );
INSERT INTO TRIP VALUES(14, 20002, 'PKR008', '2015-04-23' );
INSERT INTO TRIP VALUES(15, 30005, 'PKR768', '2015-05-24' );
INSERT INTO TRIP VALUES(16, 30005, 'SST005', '2014-08-02' );
INSERT INTO TRIP VALUES(17, 20002, 'QRT834', '2014-09-17' );
INSERT INTO TRIP VALUES(18, 10001, 'KKK007', '2014-12-15' );
INSERT INTO TRIP VALUES(19, 30005, 'SST005', '2016-01-23' );
INSERT INTO TRIP VALUES(20, 10003, 'PKR768', '2016-03-12' );
INSERT INTO TRIP VALUES(21, 10001, 'QRT834', '2012-04-23' );
INSERT INTO TRIP VALUES(22, 30005, 'PKR008', '2012-04-23' );
INSERT INTO TRIP VALUES(23, 10003, 'PKR768', '2012-05-25' );
INSERT INTO TRIP VALUES(24, 20002, 'SST005', '2012-08-02' );
INSERT INTO TRIP VALUES(25, 10001, 'PKR768', '2014-01-12' );
INSERT INTO TRIP VALUES(26, 10001, 'SYF777', '2013-02-20' );
INSERT INTO TRIP VALUES(27, 20002, 'KKK007', '2013-03-12' );
INSERT INTO TRIP VALUES(28, 30005, 'PKR768', '2010-06-29' );
INSERT INTO TRIP VALUES(29, 10001, 'QRT834', '2010-09-17' );
INSERT INTO TRIP VALUES(30, 10002, 'KKK007', '2010-12-15' );
INSERT INTO TRIP VALUES(31, 10003, 'SST005', '2010-01-23' );
INSERT INTO TRIP VALUES(32, 20002, 'PKR768', '2010-03-12' );
INSERT INTO TRIP VALUES(33, 30005, 'QRT834', '2003-04-23' );
INSERT INTO TRIP VALUES(34, 30005, 'PKR008', '2004-04-23' );
INSERT INTO TRIP VALUES(35, 10001, 'PKR768', '2017-05-24' );

INSERT INTO DRIVER VALUES( 1, 10001, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 2, 10008, 'ON LEAVE' );
INSERT INTO DRIVER VALUES( 3, 10002, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 4, 10004, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 5, 10003, 'ON LEAVE' );
INSERT INTO DRIVER VALUES( 6, 10012, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 7, 20002, 'BUSY' );
INSERT INTO DRIVER VALUES( 8, 20003, 'BUSY' );
INSERT INTO DRIVER VALUES( 9, 30005, 'BUSY' );
INSERT INTO DRIVER VALUES( 10, 40002, 'BUSY' );
INSERT INTO DRIVER VALUES( 11, 20045, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 12, 20055, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 13, 20065, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 14, 10305, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 15, 10345, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 16, 10705, 'AVAILABLE' );
INSERT INTO DRIVER VALUES( 17, 40005, 'AVAILABLE' );

INSERT INTO EMPLOYEE VALUES( 1, 'John', NULL, 'Smith', NULL, 42, 
'Victoria St.', 'Hurstville', 'NSW', 2456 );
INSERT INTO EMPLOYEE VALUES( 2, 'Peter', NULL, 'Taylor', '1970-01- 
12', 42, 'Victoria St.', 'Hurstville', 'NSW', 2456 );
INSERT INTO EMPLOYEE VALUES( 3, 'John', NULL, 'Doe', '1966-03-23', 
12, 'Station St.', 'Dapto', 'NSW', 2530 );
INSERT INTO EMPLOYEE VALUES( 4, 'John', NULL, 'Gray', '1988-05-05', 
16, 'Station St.', 'Dapto', 'NSW', 2530 );
INSERT INTO EMPLOYEE VALUES( 5, 'Adam', NULL, 'Taylor', '1980-01-01', 
42, 'Church St.', 'City', 'NSW', 2300 );
INSERT INTO EMPLOYEE VALUES( 6, 'Michael', NULL, 'Jones', '1975-03- 
05', 23, 'Waterloo Ave.', 'Surry Hills', 'NSW', 2502 );
INSERT INTO EMPLOYEE VALUES( 7, 'Frederic', NULL, 'Jones', NULL, 3, 
'Victoria St.', 'Redfern', 'NSW', 2420 );
INSERT INTO EMPLOYEE VALUES( 8, 'Peter', NULL, 'O''Brien', '1983-02- 
28', 19, 'Lucas Dr.', 'Horsley', 'NSW', 2530 );
INSERT INTO EMPLOYEE VALUES( 9, 'John', NULL, 'Lucas', '1966-12-16', 
20, 'Huxley St.', 'Horsley', 'NSW', 2530 );
INSERT INTO EMPLOYEE VALUES( 10, 'John', NULL, 'Fox', '1975-10-15', 
18, 'Victoria St.', 'Hurstville', 'NSW', 2456 );
INSERT INTO EMPLOYEE VALUES( 11, 'Adam', NULL, 'Fox', NULL, 45, 
'Victoria St.', 'Hurstville', 'NSW', 2456 );
INSERT INTO EMPLOYEE VALUES( 12, 'Phillip', NULL, ',Cox', '1984-12- 
12', 5, 'The Avenue', 'Rockdale', 'NSW', 2300 );
INSERT INTO EMPLOYEE VALUES( 13, 'Andrew', 'K', 'Smith', '1969-04- 
04', 42, 'Bambaramba Ave.', 'Pennant Hills', 'NSW', 2556 );
INSERT INTO EMPLOYEE VALUES( 14, 'Andrew', 'R', 'Smith', '1992-04- 
01', 67, 'King Cr.', 'Hurstville', 'NSW', 2456 );
INSERT INTO EMPLOYEE VALUES( 15, 'Michael', NULL, 'Potter', '1995-04- 
01', 568, 'Bong Bong St.', 'Horsley', 'NSW', 2530 );
INSERT INTO EMPLOYEE VALUES( 16, 'Harry', NULL, 'Potter', '1995-04- 
01', 568, 'Bong Bong St.', 'Horsley', 'NSW', 2530 );
INSERT INTO EMPLOYEE VALUES( 17, 'James', NULL, 'Bond', NULL, 007, 
'Alan Bond St.', 'Perth', 'WA', 6000 );
INSERT INTO EMPLOYEE VALUES( 18, 'Paris', NULL, 'Hilton', NULL, 1, 
'Hilton St.', 'Melbourne', 'Vic', 3000 );
INSERT INTO EMPLOYEE VALUES( 19, 'Lady', NULL, 'Gaga', NULL, 3, 'Pork 
st.', 'Hobart', 'Tas', 7000 );
INSERT INTO EMPLOYEE VALUES( 20, 'Robin', NULL, 'Hood', NULL, 6, 
'Nottingham Pl.', 'Sydney', 'NSW', 2000 );

但是在运行代码后，我出现此错误： （）缺少1个必需的位置参数：“ y”

所以我尝试用以下方式更改最后一行：

import numpy as np
from scipy.stats import norm
from scipy.integrate import quad
from scipy.optimize import root

c = 10

f = lambda x, y: norm.pdf(y-x)
eq = lambda x, y: quad(f, -np.inf, np.inf, args=(y))[0] - c
sol = root(eq, x0=0)

print('x =', int(sol.x[0]))

或

sol = lambda x, y: root(eq, x0=0, args=(y))

但是我不知道它是否有效，如果可以，如何获取变量中的解决方案？

非常感谢您的帮助！ ：-）