Find the value of ∫ from 0 to 2 of (x^2 - 2x + 1) dx.
Practice Questions
1 question
Q1
Find the value of ∫ from 0 to 2 of (x^2 - 2x + 1) dx.
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The integral evaluates to [x^3/3 - x^2 + x] from 0 to 2 = (8/3 - 4 + 2) = 2/3.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the value of ∫ from 0 to 2 of (x^2 - 2x + 1) dx.
Solution: The integral evaluates to [x^3/3 - x^2 + x] from 0 to 2 = (8/3 - 4 + 2) = 2/3.
Steps: 7
Step 1: Identify the function to integrate, which is (x^2 - 2x + 1).
Step 2: Find the antiderivative of the function. The antiderivative of x^2 is x^3/3, the antiderivative of -2x is -x^2, and the antiderivative of 1 is x.
Step 3: Combine the antiderivatives to get the complete antiderivative: (x^3/3 - x^2 + x).
Step 4: Evaluate the antiderivative from the lower limit (0) to the upper limit (2).
Step 5: Substitute 2 into the antiderivative: (2^3/3 - 2^2 + 2) = (8/3 - 4 + 2).
