6.] Replace the integral in exercise 5 (int. (1/ 1 – t) dt a = 0, b = 1/2with ?1/(1+t) dt with a = 0, b = 1, and repeat the four steps.

a. integrate using a graphing utility
b. integrate exactly
c. integrate by replacing the integrand with a Taylor series and integrate term by term.
d. rewrite your answer to c using summarization notation and equate it to your answer in part b.

To replace the integral in Exercise 5 with the new integrand ?1/(1+t) dt with a = 0, b = 1, we will follow the four steps:

a. Integrate using a graphing utility:
Use a graphing utility, such as Wolfram Alpha or an online graphing calculator, to evaluate the integral. Simply enter the expression ?1/(1+t) dt, with the limits of integration a = 0 and b = 1, and let the calculator compute the result. This will give you the approximate value of the integral.

b. Integrate exactly:
To integrate the given expression exactly, we can use techniques like substitution or partial fractions. In this case, integrating ?1/(1+t) dt is straightforward. Using the substitution u = 1+t, we get du = dt. When t = 0, u = 1, and when t = 1, u = 2. So the integral becomes:

∫ ?1/(1+t) dt = ∫ ?1/u du = -ln|u| = -ln|1+t| evaluated from 0 to 1.

Substituting the limits, we have:

- ln|2| - (-ln|1|) = -ln(2) - 0 = -ln(2).

Therefore, the exact value of the integral is -ln(2).

c. Integrate by replacing the integrand with a Taylor series and integrate term by term:
To integrate the given expression using a Taylor series, we expand the function ?1/(1+t) as a power series. The Taylor series expansion for this function is:

?1/(1+t) = -∑((-t)^n), where n goes from 0 to infinity.

We can integrate term by term by taking the integral of each term of the power series. The integral of (-t)^n dt is (-1/n+1)*t^(n+1). Taking the integral of the entire series, we get:

∫ ?1/(1+t) dt = -∑((-1/n+1)*t^(n+1)), where n goes from 0 to infinity.

d. Rewrite your answer to c using summation notation and equate it to your answer in part b:
Using summation notation, we can rewrite the answer from part c as:

∫ ?1/(1+t) dt = -∑((-1/n+1)*t^(n+1)), where n goes from 0 to infinity.

To equate this to the exact value found in part b, we can sum up the terms and set it equal to -ln(2). The series expansion for ln(2) is given by:

ln(2) = -∑((-1)^n)/(n+1), where n goes from 0 to infinity.

Equating the two equations, we have:

-∑((-1/n+1)*t^(n+1)) = -∑((-1)^n)/(n+1).

This allows us to relate the Taylor series representation of the integral to the infinite series representation of ln(2).