Find the value of ∫ from 0 to 1 of (x^3 - 4x + 4) dx.

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Question: Find the value of ∫ from 0 to 1 of (x^3 - 4x + 4) dx.

Options:

Correct Answer: 1

Solution:

The integral evaluates to [x^4/4 - 2x^2 + 4x] from 0 to 1 = (1/4 - 2 + 4) = (1/4 + 2) = 9/4.

Find the value of ∫ from 0 to 1 of (x^3 - 4x + 4) dx.

Find the value of ∫ from 0 to 1 of (x^3 - 4x + 4) dx.

Step 1: Write down the integral you need to solve: ∫ from 0 to 1 of (x^3 - 4x + 4) dx.
Step 2: Find the antiderivative of the function (x^3 - 4x + 4).
Step 3: The antiderivative is calculated as follows: For x^3, the antiderivative is (x^4)/4; for -4x, it is -2x^2; and for +4, it is 4x.
Step 4: Combine these results to get the complete antiderivative: (x^4)/4 - 2x^2 + 4x.
Step 5: Now evaluate this antiderivative from 0 to 1. First, substitute 1 into the antiderivative: (1^4)/4 - 2(1^2) + 4(1).
Step 6: Calculate the value: (1/4) - 2 + 4 = (1/4) + 2.
Step 7: Convert 2 into a fraction with a denominator of 4: 2 = 8/4, so (1/4) + (8/4) = 9/4.
Step 8: Finally, write down the result: The value of the integral is 9/4.

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