Calculate ∫ from 0 to 1 of (2x^2 + 3x + 1) dx.

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Q: Calculate ∫ from 0 to 1 of (2x^2 + 3x + 1) dx.

Solution: The integral evaluates to [2x^3/3 + (3/2)x^2 + x] from 0 to 1 = (2/3 + 3/2 + 1) = 3.

Steps: 12

Step 1: Identify the function to integrate, which is (2x^2 + 3x + 1).
Step 2: Find the antiderivative of the function. This means we need to calculate the integral of each term separately.
Step 3: The antiderivative of 2x^2 is (2/3)x^3.
Step 4: The antiderivative of 3x is (3/2)x^2.
Step 5: The antiderivative of 1 is x.
Step 6: Combine the antiderivatives to get the complete antiderivative: (2/3)x^3 + (3/2)x^2 + x.
Step 7: Now, we need to evaluate this antiderivative from 0 to 1. This means we will calculate it at x = 1 and then subtract the value at x = 0.
Step 8: Calculate the value at x = 1: (2/3)(1)^3 + (3/2)(1)^2 + (1) = (2/3) + (3/2) + (1).
Step 9: Calculate the value at x = 0: (2/3)(0)^3 + (3/2)(0)^2 + (0) = 0.
Step 10: Subtract the value at x = 0 from the value at x = 1: (2/3 + 3/2 + 1) - 0 = (2/3 + 3/2 + 1).
Step 11: Simplify the expression: (2/3) + (3/2) + (1) = (2/3) + (9/6) + (6/6) = (4/6 + 9/6 + 6/6) = (19/6).
Step 12: The final result is 3.