Find the value of ∫ from 0 to 1 of (x^4 - 4x^3 + 6x^2 - 4x + 1) dx.
Practice Questions
1 question
Q1
Find the value of ∫ from 0 to 1 of (x^4 - 4x^3 + 6x^2 - 4x + 1) dx.
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The integral evaluates to [x^5/5 - x^4 + 2x^3 - 2x^2 + x] from 0 to 1 = (1/5 - 1 + 2 - 2 + 1) = 0.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the value of ∫ from 0 to 1 of (x^4 - 4x^3 + 6x^2 - 4x + 1) dx.
Solution: The integral evaluates to [x^5/5 - x^4 + 2x^3 - 2x^2 + x] from 0 to 1 = (1/5 - 1 + 2 - 2 + 1) = 0.
Steps: 10
Step 1: Identify the function to integrate, which is f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1.
Step 2: Find the antiderivative (indefinite integral) of f(x). This means we need to integrate each term separately.
Step 3: Integrate each term: The integral of x^4 is (x^5)/5, the integral of -4x^3 is -x^4, the integral of 6x^2 is 2x^3, the integral of -4x is -2x^2, and the integral of 1 is x.
Step 4: Combine the results of the integrations: The antiderivative is (x^5)/5 - x^4 + 2x^3 - 2x^2 + x.
Step 5: Evaluate the definite integral from 0 to 1. This means we will calculate the antiderivative at 1 and subtract the value of the antiderivative at 0.
Step 6: Calculate the antiderivative at x = 1: (1^5)/5 - (1^4) + 2(1^3) - 2(1^2) + 1 = 1/5 - 1 + 2 - 2 + 1.
