Compute Integrals: Integrate (z^3-6z^2+4)dz from -1+i to 1

[asqw121]

Integrate (z^3-6z^2+4)dz where the function is any curve joining -1+i to 1. Z is complex number

 

[MarkFL]

Please post your progress so far so our helpers know exactly where you are stuck or what you may be doing wrong. (Nod)

 

[asqw121]

I am trying to first separate them to integrate individual z^3dz, -6z^2dz and 4dz.

Line integral γ(t)= t*i+(1-t)*(-1-+i)

 

[alyafey22]

There is no need to find a parametric representation of the curve because the function you are integrating has an anti derivative , indeed it is a polynomial . By independence of the path , the integral only depends on the initial and final points .

 

[blue_raver22]

I would approach this problem using the fundamental theorem of calculus, which states that the integral of a function can be calculated by finding its antiderivative and evaluating it at the endpoints of the integration interval.

For the first part of the problem, integrating (z^3-6z^2+4)dz from -1+i to 1 would involve finding the antiderivative of the function, which is z^4/4 - 2z^3 + 4z. Then, we would plug in the endpoints (-1+i and 1) into this antiderivative and subtract the result at the lower endpoint from the result at the upper endpoint. This would give us the value of the integral.

For the second part of the problem, where the function is any curve joining -1+i to 1, we would need to use the line integral. This involves parameterizing the curve and then integrating the function along the curve using the parameter as the integration variable. The result would be a complex number representing the integral along the given curve.

In both cases, the value of the integral would depend on the specific path or curve chosen. It would be helpful to plot the function and the given points to visualize the curve and better understand the integration process. Additionally, for complex numbers, it is important to consider the orientation of the curve and whether it is a closed or open curve, as this can affect the value of the integral.