What is the value of the integral ∫(2x^3 - 3x^2 + 4)dx from 1 to 2?

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Q: What is the value of the integral ∫(2x^3 - 3x^2 + 4)dx from 1 to 2?

Solution: The integral evaluates to [0.5x^4 - x^3 + 4x] from 1 to 2 = (8 - 8 + 8) - (0.5 - 1 + 4) = 8 - 2.5 = 5.5.

Steps: 13

Step 1: Identify the function to integrate, which is 2x^3 - 3x^2 + 4.
Step 2: Find the antiderivative (indefinite integral) of the function. This means we need to integrate each term separately.
Step 3: The antiderivative of 2x^3 is (2/4)x^4 = 0.5x^4.
Step 4: The antiderivative of -3x^2 is (-3/3)x^3 = -x^3.
Step 5: The antiderivative of 4 is 4x.
Step 6: Combine the antiderivatives to get the complete antiderivative: 0.5x^4 - x^3 + 4x.
Step 7: Now, evaluate this antiderivative from the lower limit (1) to the upper limit (2).
Step 8: First, substitute x = 2 into the antiderivative: 0.5(2^4) - (2^3) + 4(2).
Step 9: Calculate: 0.5(16) - 8 + 8 = 8 - 8 + 8 = 8.
Step 10: Next, substitute x = 1 into the antiderivative: 0.5(1^4) - (1^3) + 4(1).
Step 11: Calculate: 0.5(1) - 1 + 4 = 0.5 - 1 + 4 = 3.5.
Step 12: Now, subtract the value at the lower limit from the value at the upper limit: 8 - 3.5.
Step 13: Calculate the final result: 8 - 3.5 = 4.5.