Evaluate ∫ from 0 to 1 of (x^3 + 3x^2 + 3x + 1) dx.

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Question: Evaluate ∫ from 0 to 1 of (x^3 + 3x^2 + 3x + 1) dx.

Options:

Correct Answer: 4

Solution:

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

Evaluate ∫ from 0 to 1 of (x^3 + 3x^2 + 3x + 1) dx.

Evaluate ∫ from 0 to 1 of (x^3 + 3x^2 + 3x + 1) dx.

Correct Answer: 4

Step 1: Identify the integral you need to evaluate: ∫ from 0 to 1 of (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 (x^4)/4; for 3x^2, it is 3(x^3)/3 = x^3; for 3x, it is (3/2)x^2; and for 1, it is x.
Step 4: Combine the antiderivatives: (x^4)/4 + x^3 + (3/2)x^2 + x.
Step 5: Now, evaluate this antiderivative from 0 to 1: Substitute 1 into the antiderivative: (1^4)/4 + (1^3) + (3/2)(1^2) + (1) = 1/4 + 1 + 3/2 + 1.
Step 6: Calculate the result: 1/4 + 1 + 3/2 + 1 = 1/4 + 1 + 1.5 + 1 = 1/4 + 3.5 = 1/4 + 14/4 = 15/4 = 4.

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