4435.sagews

# integrate e^z from 0 to 1 to 1+i

f(z) = e^z # define the integrand

c1(t) = t # parametrized where c1(t):[0,1]
c2(t) = 1+I*t # parametrized where c2(t):[0,1]

start = 0
stop = 1
N = 1000

line_points = srange(float(start), float(stop), (stop-start)/N,include_endpoint=True)  

z1=map(c1,line_points)
z2=map(c2,line_points)

X = sum(f(z1[i])*(z1[i+1]-z1[i]) for i in range(0,N-1))
Y = sum(f(z2[i])*(z2[i+1]-z2[i]) for i in range(0,N-1))
X+Y # Approximation of the integral sums

0.468549044512 + 2.28650890995*I

dc1(t)=c1.diff(t)    # derivative of c1(t)
dc2(t)=c2.diff(t)    # derivative of c2(t)
answer1=integrate(f(c1(t))*dc1(t),(t,start,stop))
answer2=integrate(f(c2(t))*dc2(t),(t,start,stop))
answer=answer1+answer2
answer # final solution of adding the two integrals

e^(I + 1) - 1

float(answer.real_part())+float(answer.imag_part())*I
    # check if answer from parametrization agrees with approximation by taking its floating-point value

0.468693939916 + 2.28735528718*I

# Now as we can see, the approximation sums and parametrization solutions are very close