Find the value of ∫ from 0 to 1 of (x^2 + 3x + 2) dx.
Practice Questions
1 question
Q1
Find the value of ∫ from 0 to 1 of (x^2 + 3x + 2) dx.
1
2
3
4
The integral evaluates to [x^3/3 + (3/2)x^2 + 2x] from 0 to 1 = (1/3 + 3/2 + 2) - 0 = 4/3 + 3/2 = 2.5.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the value of ∫ 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) - 0 = 4/3 + 3/2 = 2.5.
Steps: 9
Step 1: Identify the function to integrate, which is (x^2 + 3x + 2).
Step 2: Find the antiderivative of the function. The antiderivative of x^2 is (x^3)/3, the antiderivative of 3x is (3/2)x^2, and the antiderivative of 2 is 2x.
Step 3: Combine the antiderivatives to get the complete antiderivative: (x^3)/3 + (3/2)x^2 + 2x.
Step 4: Evaluate the antiderivative from 0 to 1. This means you will calculate the value of the antiderivative at 1 and then subtract the value at 0.
Step 5: Calculate the value at 1: (1^3)/3 + (3/2)(1^2) + 2(1) = (1/3) + (3/2) + 2.
Step 6: Calculate the value at 0: (0^3)/3 + (3/2)(0^2) + 2(0) = 0.
Step 7: Subtract the value at 0 from the value at 1: ((1/3) + (3/2) + 2) - 0.
