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

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

Options:

Correct Answer: 1

Solution:

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

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

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

Step 1: Identify the function to integrate, which is (x^3 - 3x^2 + 2).
Step 2: Find the antiderivative of the function. This means we need to integrate each term separately.
Step 3: The antiderivative of x^3 is (x^4)/4.
Step 4: The antiderivative of -3x^2 is -3(x^3)/3 = -x^3.
Step 5: The antiderivative of 2 is 2x.
Step 6: Combine the antiderivatives to get the complete antiderivative: (x^4)/4 - x^3 + 2x.
Step 7: Now, evaluate this antiderivative from 0 to 1. This means we will calculate it at 1 and then subtract the value at 0.
Step 8: Calculate the value at 1: (1^4)/4 - (1^3) + 2(1) = 1/4 - 1 + 2.
Step 9: Simplify the expression: 1/4 - 1 + 2 = 1/4 + 1 = 5/4.
Step 10: Calculate the value at 0: (0^4)/4 - (0^3) + 2(0) = 0.
Step 11: Subtract the value at 0 from the value at 1: (5/4) - (0) = 5/4.
Step 12: The final answer is 5/4.

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