The integral evaluates to [2x^3 - 2x^2 + x] from 0 to 1 = (2 - 2 + 1) = 1.
Questions & Step-by-step Solutions
1 item
Q
Q: Calculate ∫ from 0 to 1 of (6x^2 - 4x + 1) dx.
Solution: The integral evaluates to [2x^3 - 2x^2 + x] from 0 to 1 = (2 - 2 + 1) = 1.
Steps: 7
Step 1: Identify the function to integrate, which is (6x^2 - 4x + 1).
Step 2: Find the antiderivative of the function. The antiderivative of 6x^2 is 2x^3, the antiderivative of -4x is -2x^2, and the antiderivative of 1 is x.
Step 3: Combine the antiderivatives to get the complete antiderivative: 2x^3 - 2x^2 + x.
Step 4: Evaluate the antiderivative from the lower limit (0) to the upper limit (1).
Step 5: Substitute the upper limit (1) into the antiderivative: 2(1)^3 - 2(1)^2 + (1) = 2 - 2 + 1 = 1.
Step 6: Substitute the lower limit (0) into the antiderivative: 2(0)^3 - 2(0)^2 + (0) = 0.
Step 7: Subtract the value from the lower limit from the value from the upper limit: 1 - 0 = 1.
