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What is the value of the integral ∫_0^1 (3x^2 + 2x) dx?
What is the value of the integral ∫_0^1 (3x^2 + 2x) dx?
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Q1
What is the value of the integral ∫_0^1 (3x^2 + 2x) dx?
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Evaluating the integral gives [x^3 + x^2]_0^1 = (1 + 1) - (0 + 0) = 2.
Questions & Step-by-step Solutions
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Q
Q: What is the value of the integral ∫_0^1 (3x^2 + 2x) dx?
Solution:
Evaluating the integral gives [x^3 + x^2]_0^1 = (1 + 1) - (0 + 0) = 2.
Steps: 6
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Step 1: Identify the integral to evaluate, which is ∫_0^1 (3x^2 + 2x) dx.
Step 2: Find the antiderivative of the function 3x^2 + 2x. The antiderivative is x^3 + x^2.
Step 3: Use the Fundamental Theorem of Calculus to evaluate the antiderivative from 0 to 1. This means we will calculate [x^3 + x^2] from 0 to 1.
Step 4: Substitute the upper limit (1) into the antiderivative: (1^3 + 1^2) = 1 + 1 = 2.
Step 5: Substitute the lower limit (0) into the antiderivative: (0^3 + 0^2) = 0 + 0 = 0.
Step 6: Subtract the result from the lower limit from the result from the upper limit: 2 - 0 = 2.
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