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

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

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: Integrate 2x^3. The antiderivative is (2/4)x^4 = (1/2)x^4.
Step 4: Integrate -3x^2. The antiderivative is (-3/3)x^3 = -x^3.
Step 5: Integrate 4. The antiderivative is 4x.
Step 6: Combine the antiderivatives. The complete antiderivative is (1/2)x^4 - x^3 + 4x + C, where C is the constant of integration.
Step 7: Evaluate the definite integral from 1 to 3. This means we will calculate the antiderivative at 3 and subtract the value at 1.
Step 8: Calculate the value at x = 3: (1/2)(3^4) - (3^3) + 4(3) = (1/2)(81) - 27 + 12 = 40.5 - 27 + 12 = 25.5.
Step 9: Calculate the value at x = 1: (1/2)(1^4) - (1^3) + 4(1) = (1/2)(1) - 1 + 4 = 0.5 - 1 + 4 = 3.5.
Step 10: Subtract the two results: 25.5 - 3.5 = 22.
Step 11: The final answer is 22.

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