Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x - 1) dx.

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Question: Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x - 1) dx.

Options:

Correct Answer: 0

Solution:

The integral evaluates to [x^4/4 - x^3 + (3/2)x^2 - x] from 0 to 1 = 0.

Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x - 1) dx.

Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x - 1) dx.

Correct Answer: 0

Step 1: Identify the function to integrate, which is f(x) = x^3 - 3x^2 + 3x - 1.
Step 2: Find the antiderivative (indefinite integral) of f(x). This means we need to integrate each term separately.
Step 3: Integrate x^3 to get (1/4)x^4.
Step 4: Integrate -3x^2 to get -x^3.
Step 5: Integrate 3x to get (3/2)x^2.
Step 6: Integrate -1 to get -x.
Step 7: Combine all the results from Steps 3 to 6 to get the antiderivative: (1/4)x^4 - x^3 + (3/2)x^2 - x.
Step 8: Now evaluate the antiderivative from 0 to 1. This means we will calculate the value at x = 1 and then subtract the value at x = 0.
Step 9: Calculate the value at x = 1: (1/4)(1)^4 - (1)(1)^3 + (3/2)(1)^2 - (1) = 1/4 - 1 + 3/2 - 1 = 0.
Step 10: Calculate the value at x = 0: (1/4)(0)^4 - (1)(0)^3 + (3/2)(0)^2 - (0) = 0.
Step 11: Subtract the value at x = 0 from the value at x = 1: 0 - 0 = 0.
Step 12: Therefore, the value of the integral from 0 to 1 is 0.

Definite Integral – The question tests the ability to evaluate a definite integral of a polynomial function over a specified interval.
Polynomial Integration – It assesses knowledge of integrating polynomial functions and applying the Fundamental Theorem of Calculus.