The integral evaluates to [x^3/3 + x^2 + x] from 0 to 1 = (1/3 + 1 + 1) - (0) = 7/3.
Questions & Step-by-step Solutions
1 item
Q
Q: Calculate ∫ from 0 to 1 of (x^2 + 2x + 1) dx.
Solution: The integral evaluates to [x^3/3 + x^2 + x] from 0 to 1 = (1/3 + 1 + 1) - (0) = 7/3.
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 from the lower limit (0) to the upper limit (1).
Step 5: Substitute the upper limit (1) into the antiderivative: (1^3/3 + 1^2 + 1) = (1/3 + 1 + 1).
Step 6: Calculate the result from the upper limit: 1/3 + 1 + 1 = 1/3 + 3/3 = 4/3.
Step 7: Substitute the lower limit (0) into the antiderivative: (0^3/3 + 0^2 + 0) = 0.
Step 8: Subtract the result of the lower limit from the result of the upper limit: (4/3 - 0) = 4/3.
