The integral evaluates to [x^3/3 + x^2 + x] from 0 to 2 = (8/3 + 4 + 2) = 6.
Questions & Step-by-step Solutions
1 item
Q
Q: Evaluate ∫ from 0 to 2 of (x^2 + 2x + 1) dx.
Solution: The integral evaluates to [x^3/3 + x^2 + x] from 0 to 2 = (8/3 + 4 + 2) = 6.
Steps: 8
Step 1: Identify the function to integrate, which is (x^2 + 2x + 1).
Step 2: Find the antiderivative of the function. The antiderivative of x^2 is x^3/3, the antiderivative of 2x is x^2, and the antiderivative of 1 is x.
Step 3: Combine the antiderivatives to get the complete antiderivative: (x^3/3 + x^2 + x).
Step 4: Evaluate the antiderivative at the upper limit (2) and the lower limit (0).
Step 5: Calculate the value at the upper limit: (2^3/3 + 2^2 + 2) = (8/3 + 4 + 2).
Step 6: Calculate the value at the lower limit: (0^3/3 + 0^2 + 0) = 0.
Step 7: Subtract the lower limit value from the upper limit value: (8/3 + 4 + 2) - 0 = (8/3 + 4 + 2).
Step 8: Simplify the expression: 4 + 2 = 6, so the final result is 6.
