Find the value of ∫ from 1 to 2 of (3x^2 - 2x + 1) dx.
Practice Questions
1 question
Q1
Find the value of ∫ from 1 to 2 of (3x^2 - 2x + 1) dx.
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The integral evaluates to [x^3 - x^2 + x] from 1 to 2 = (8 - 4 + 2) - (1 - 1 + 1) = 5.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the value of ∫ from 1 to 2 of (3x^2 - 2x + 1) dx.
Solution: The integral evaluates to [x^3 - x^2 + x] from 1 to 2 = (8 - 4 + 2) - (1 - 1 + 1) = 5.
Steps: 6
Step 1: Identify the function to integrate, which is (3x^2 - 2x + 1).
Step 2: Find the antiderivative of the function. The antiderivative of 3x^2 is x^3, the antiderivative of -2x is -x^2, and the antiderivative of 1 is x. So, the antiderivative is x^3 - x^2 + x.
Step 3: Write the definite integral from 1 to 2 using the antiderivative: [x^3 - x^2 + x] from 1 to 2.
Step 4: Calculate the value of the antiderivative at the upper limit (x = 2): 2^3 - 2^2 + 2 = 8 - 4 + 2 = 6.
Step 5: Calculate the value of the antiderivative at the lower limit (x = 1): 1^3 - 1^2 + 1 = 1 - 1 + 1 = 1.
Step 6: Subtract the lower limit result from the upper limit result: 6 - 1 = 5.
