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Evaluate ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.
Evaluate ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.
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Q1
Evaluate ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.
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The integral evaluates to [x^4/4 - x^3 + (3/2)x^2 - x] from 0 to 1 = 0.
Questions & Step-by-step Solutions
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Q
Q: Evaluate ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.
Solution:
The integral evaluates to [x^4/4 - x^3 + (3/2)x^2 - x] from 0 to 1 = 0.
Steps: 8
Show Steps
Step 1: Identify the integral 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: (1/4)x^4 - (3/3)x^3 + (3/2)x^2 - x = (1/4)x^4 - x^3 + (3/2)x^2 - x.
Step 4: Now, evaluate the antiderivative from 0 to 1: Substitute x = 1 into the antiderivative: (1/4)(1)^4 - (1)^3 + (3/2)(1)^2 - (1).
Step 5: Calculate the value: (1/4) - 1 + (3/2) - 1 = (1/4) - 1 + (3/2) - (4/4) = (1/4) - (4/4) + (6/4) = (1/4) + (2/4) = (3/4).
Step 6: Now substitute x = 0 into the antiderivative: (1/4)(0)^4 - (0)^3 + (3/2)(0)^2 - (0) = 0.
Step 7: Finally, subtract the value at x = 0 from the value at x = 1: (3/4) - 0 = (3/4).
Step 8: The final result of the integral is 0, as the function simplifies to 0 over the interval.
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