Area under y=(4/3)sqrt(9-x^2), above x-axis, and to the left of y=2x

from sympy import *

x=symbols('x')
f=4/3*sqrt(9-x**2)
g=2*x

matplotlib notebook

plot((f,(x,-3,3)),(g,(x,-3,3)))

c=solve(f-g,x)[0]
# this case only has one solution so adding [0] gives that single solution as a number, instead of a list
print(c)
Area1=integrate(f,(x,-3,0))
Area2=integrate(f-g,(x,0,c))
print(Area1,Area2)
print('The area is equal to', (Area1+Area2).evalf())

from sympy import *

x=symbols('x')
a,b=symbols('a b')
f=b/a*sqrt(a**2-x**2)

fplot=f.subs({a:3,b:4})
print(fplot)

matplotlib notebook

plot(fplot,(x,-3,3))

c=solve(f,x)
print(c)

Volume=integrate(pi*f**2,(x,c[0],c[1]))
print('The volume of the ellipsoid is',Volume.simplify())

10 Joules of work to move a spring from a natural length of 10cm to a length of 15cm

How much work would it take to move the spring from a length of 15cm to a length of 20cm?

from sympy import *

x,k=symbols('x k')
F=k*x
Work1=integrate(F,(x,0,0.05))
ksol=solve(Work1-10,k)[0]
print(ksol)
Fnew=F.subs(k,ksol)
print(Fnew)
Work=integrate(Fnew,(x,.05,.10))
print('The work required is',Work,'J')

Work required to pump water from a 0.5m spout at the top of a tank 6m long whose ends are semicircles with radius 2m

from sympy import *

y=symbols('y')
dW=9800*6*(2*sqrt(4-y**2))*(y+0.5)
W=integrate(dW,(y,0,2))
print('The work required is',W.evalf(),'Joules')

Limit of a_n = n * sin(pi/n)
a) List the first 20 terms (numerical estimate)
b) Plot the first 50 terms (graphical estimate)
c) Find the exact limit

from sympy import *

n=symbols('n',positive=True,integer=True)
a=n*sin(pi/n)
a1to20=[a.subs(n,i).evalf() for i in range(1,21)]
print(a1to20)

matplotlib notebook

import matplotlib.pyplot as plt
n1to50=range(1,51)
a1to50=[a.subs(n,i) for i in n1to50]
plt.plot(n1to50,a1to50,'o')

L=limit(a,n,oo)
print('The limit of the sequence is',L)

Given the series sum from 1 to infinity of (-1)^(n+1)/n^2
a) Find and plot the first 20 partial sums
b) Find the exact sum of the series

from sympy import *
import numpy as np

n=symbols('n',positive=True,integer=True)
a=(-1)**(n+1)/n**2
n1to20=range(1,21)
a1to20=[a.subs(n,i).evalf() for i in n1to20]
print(a1to20)
s1to20=np.cumsum(a1to20)
print(s1to20)

matplotlib notebook

import matplotlib.pyplot as plt
plt.plot(n1to20,s1to20,'o')

from mpmath import nsum,inf
S=nsum(lambda n:(-1)**(n+1)/n**2,[1,inf])
print('The sum of the series is',S)

Generate and print the first 50 Fibonacci numbers

a=[1,1]
for n in range(2,50): 
    a.append(a[n-2]+a[n-1])
print('The first 50 fibonnaci numbers are',a)

Find the radius and interval of convergence of the sum from 0 to infinity of (-1)^n * x^n/( (n+2) * 2^n)

from sympy import *

n=symbols('n',positive=True,integer=True)
x=symbols('x')
a=(-1)**n*x**n/((n+2)*2**n)
RatioTest=abs(a.subs(n,n+1)/a)
print(RatioTest.simplify())
L=limit(RatioTest.simplify(),n,oo)
print('The limit of the Ratio Test is',L) 
# likely obvious x is between -2 and 2
reduce_inequalities(L<1)

# Left endpoint
aLeft=a.subs(x,-2)
print(aLeft.simplify(),'This series diverges by limit comparison with the series 1/n.')

# Right endpoint
aRight=a.subs(x,2)
print(aRight.simplify(),'This series converges by the Alternating Series Test.')

Plot f(x)=arctan(x) and the 3rd, 5th, and 10th partial sum of its powers series

sum from 0 to infinity (-1)^n x^(2n+1) / (2n+1)

from sympy import *

n=symbols('n',positive=True,integer=True)
x=symbols('x')
a=(-1)**n*x**(2*n+1)/(2*n+1)
f=atan(x)
a1to3=[a.subs(n,i) for i in range(4)]
s3=sum(a1to3)
print(s3)
a1to5=[a.subs(n,i) for i in range(6)]
s5=sum(a1to5)
a1to10=[a.subs(n,i) for i in range(11)]
s10=sum(a1to10)

matplotlib notebook

plotf=plot((f,(x,-1.5,1.5)),(s3,(x,-1.5,1.5)),(s5,(x,-1.5,1.5)),(s10,(x,-1.5,1.5)),ylim=[-1,1])
plotf[1].line_color='r'
plotf[2].line_color='y'
plotf[3].line_color='k'
plotf.show()