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Evaluate ∫ from 0 to 1 of (x^2 + 3x + 2) dx.
Evaluate ∫ from 0 to 1 of (x^2 + 3x + 2) dx.
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Practice Questions
1 question
Q1
Evaluate ∫ from 0 to 1 of (x^2 + 3x + 2) dx.
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The integral evaluates to [x^3/3 + (3/2)x^2 + 2x] from 0 to 1 = (1/3 + 3/2 + 2) = (1/3 + 3/2 + 6/3) = 27/6 = 4.5.
Questions & Step-by-step Solutions
1 item
Q
Q: Evaluate ∫ from 0 to 1 of (x^2 + 3x + 2) dx.
Solution:
The integral evaluates to [x^3/3 + (3/2)x^2 + 2x] from 0 to 1 = (1/3 + 3/2 + 2) = (1/3 + 3/2 + 6/3) = 27/6 = 4.5.
Steps: 11
Show Steps
Step 1: Identify the function to integrate, which is f(x) = x^2 + 3x + 2.
Step 2: Find the antiderivative of f(x). This means we need to integrate each term separately.
Step 3: The integral of x^2 is (1/3)x^3.
Step 4: The integral of 3x is (3/2)x^2.
Step 5: The integral of 2 is 2x.
Step 6: Combine the results from Steps 3, 4, and 5 to get the antiderivative: F(x) = (1/3)x^3 + (3/2)x^2 + 2x.
Step 7: Evaluate the antiderivative from 0 to 1. This means we will calculate F(1) - F(0).
Step 8: Calculate F(1): F(1) = (1/3)(1)^3 + (3/2)(1)^2 + 2(1) = 1/3 + 3/2 + 2.
Step 9: Convert 2 into a fraction with a common denominator: 2 = 6/3.
Step 10: Now add the fractions: 1/3 + 3/2 + 6/3 = 1/3 + 4.5 = 27/6.
Step 11: Simplify the result: 27/6 = 4.5.
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