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What is the value of the integral ∫(2x + 3)dx from 0 to 1?
What is the value of the integral ∫(2x + 3)dx from 0 to 1?
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What is the value of the integral ∫(2x + 3)dx from 0 to 1?
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The integral ∫(2x + 3)dx = [x^2 + 3x] from 0 to 1 = (1^2 + 3*1) - (0 + 0) = 1 + 3 = 4.
Questions & Step-by-step Solutions
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Q: What is the value of the integral ∫(2x + 3)dx from 0 to 1?
Solution:
The integral ∫(2x + 3)dx = [x^2 + 3x] from 0 to 1 = (1^2 + 3*1) - (0 + 0) = 1 + 3 = 4.
Steps: 7
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Step 1: Identify the function to integrate, which is (2x + 3).
Step 2: Find the antiderivative of (2x + 3). The antiderivative is x^2 + 3x.
Step 3: Set up the definite integral from 0 to 1: [x^2 + 3x] from 0 to 1.
Step 4: Calculate the value of the antiderivative at the upper limit (1): (1^2 + 3*1) = 1 + 3 = 4.
Step 5: Calculate the value of the antiderivative at the lower limit (0): (0^2 + 3*0) = 0 + 0 = 0.
Step 6: Subtract the lower limit result from the upper limit result: 4 - 0 = 4.
Step 7: The final value of the integral is 4.
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