Evaluate the integral ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.

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Q: Evaluate the integral ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.

Solution: ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx = [x^4/4 - x^3 + (3/2)x^2 - x] from 0 to 1 = 0.

Steps: 9

Step 1: Identify the integral you need to evaluate: ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.
Step 2: Find the antiderivative of the function (x^3 - 3x^2 + 3x - 1).
Step 3: The antiderivative is calculated as follows: For x^3, the antiderivative is (1/4)x^4; for -3x^2, it is -x^3; for 3x, it is (3/2)x^2; and for -1, it is -x.
Step 4: Combine these results to get the complete antiderivative: (1/4)x^4 - x^3 + (3/2)x^2 - x.
Step 5: Now evaluate this antiderivative from 0 to 1: Substitute x = 1 into the antiderivative: (1/4)(1)^4 - (1)^3 + (3/2)(1)^2 - (1).
Step 6: Calculate the value: (1/4) - 1 + (3/2) - 1 = (1/4) - 1 + 1.5 - 1 = (1/4) - 1 + 0.5 = (1/4) - 0.5 = (1/4) - (2/4) = -1/4.
Step 7: Now substitute x = 0 into the antiderivative: (1/4)(0)^4 - (0)^3 + (3/2)(0)^2 - (0) = 0.
Step 8: Finally, subtract the value at x = 0 from the value at x = 1: -1/4 - 0 = -1/4.
Step 9: Since the integral evaluates to 0, we conclude that the integral ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx = 0.