Evaluate ∫ from 0 to 1 of (4x^3 - 3x^2 + 2) dx.

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Question: Evaluate ∫ from 0 to 1 of (4x^3 - 3x^2 + 2) dx.

Options:

Correct Answer: 3

Solution:

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

Evaluate ∫ from 0 to 1 of (4x^3 - 3x^2 + 2) dx.

Evaluate ∫ from 0 to 1 of (4x^3 - 3x^2 + 2) dx.

Step 1: Identify the integral you need to evaluate: ∫ from 0 to 1 of (4x^3 - 3x^2 + 2) dx.
Step 2: Find the antiderivative of the function (4x^3 - 3x^2 + 2).
Step 3: The antiderivative is calculated as follows: For 4x^3, the antiderivative is (4/4)x^4 = x^4; for -3x^2, it is (-3/3)x^3 = -x^3; and for 2, it is 2x.
Step 4: Combine these results to get the complete antiderivative: x^4 - x^3 + 2x.
Step 5: Now evaluate this antiderivative from 0 to 1: Substitute 1 into the antiderivative: (1^4 - 1^3 + 2*1) = (1 - 1 + 2) = 2.
Step 6: Substitute 0 into the antiderivative: (0^4 - 0^3 + 2*0) = (0 - 0 + 0) = 0.
Step 7: Finally, subtract the value at the lower limit from the value at the upper limit: 2 - 0 = 2.

Definite Integral – The process of calculating the area under a curve defined by a function over a specific interval.
Fundamental Theorem of Calculus – Relates differentiation and integration, allowing the evaluation of definite integrals using antiderivatives.
Polynomial Integration – Involves finding the antiderivative of polynomial functions, which is straightforward but requires careful application of power rules.