Find the value of the definite integral ∫(1 to 3) (x^2 - 2x + 1) dx. (2021)

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Question: Find the value of the definite integral ∫(1 to 3) (x^2 - 2x + 1) dx. (2021)

Options:

Correct Answer: 0

Exam Year: 2021

Solution:

∫(1 to 3) (x^2 - 2x + 1) dx = [(x^3/3 - x^2 + x)] from 1 to 3 = (9/3 - 9 + 3) - (1/3 - 1 + 1) = 0.

Find the value of the definite integral ∫(1 to 3) (x^2 - 2x + 1) dx. (2021)

Find the value of the definite integral ∫(1 to 3) (x^2 - 2x + 1) dx. (2021)

Step 1: Identify the function to integrate, which is f(x) = x^2 - 2x + 1.
Step 2: Find the antiderivative of f(x). The antiderivative is F(x) = (x^3/3) - (2x^2/2) + (1x) = (x^3/3) - x^2 + x.
Step 3: Evaluate the antiderivative at the upper limit (x = 3). Calculate F(3) = (3^3/3) - (3^2) + (3) = (27/3) - 9 + 3 = 9 - 9 + 3 = 3.
Step 4: Evaluate the antiderivative at the lower limit (x = 1). Calculate F(1) = (1^3/3) - (1^2) + (1) = (1/3) - 1 + 1 = (1/3) - 1 + 1 = (1/3).
Step 5: Subtract the value at the lower limit from the value at the upper limit: F(3) - F(1) = 3 - (1/3).
Step 6: Convert 3 to a fraction with a denominator of 3: 3 = 9/3. Now calculate: 9/3 - 1/3 = 8/3.
Step 7: The final answer is 8/3.

Definite Integral – The process of calculating the area under a curve defined by a function over a specific interval.
Polynomial Functions – Understanding how to integrate polynomial expressions, which involves applying the power rule.
Fundamental Theorem of Calculus – Using the theorem to evaluate the definite integral by finding the antiderivative and applying the limits.