Answer:
Therefore, the value of the definite integral ∫[1 to 6] √(1/x) √(16x² - 7) dx is 4√(569) - 2.
Explanation:
1. Simplify the expression:
The square root of 1/x can be written as x^(-1/2), and the square root of 16x² - 7 can be simplified as well.
∫[1 to 6] x^(-1/2) √(16x² - 7) dx
2. Apply the power rule for integration:
Using the power rule, we can integrate each term separately. The integral of x^n is (x^(n+1))/(n+1).
∫[1 to 6] x^(-1/2) √(16x² - 7) dx = [2/3 * x^(1/2) * √(16x² - 7)] [1 to 6]
3. Evaluate the integral limits:
Substituting the upper limit (6) and the lower limit (1) into the integral expression:
[2/3 * 6^(1/2) * √(16(6)² - 7)] - [2/3 * 1^(1/2) * √(16(1)² - 7)]
4. Simplify the expression:
Calculate the values within the square roots and evaluate the square roots themselves:
[2/3 * 6^(1/2) * √(576 - 7)] - [2/3 * 1^(1/2) * √(16 - 7)]
[2/3 * 6^(1/2) * √(569)] - [2/3 * 1^(1/2) * √(9)]
5. Further simplify and calculate the result:
Evaluate the square roots and simplify the expression:
[2/3 * 6^(1/2) * √(569)] - [2/3 * 1^(1/2) * 3]
[2/3 * √(36) * √(569)] - [2/3 * 3]
[2/3 * 6 * √(569)] - [2]
12/3 * √(569) - 2
4√(569) - 2
Therefore, the value of the definite integral ∫[1 to 6] √(1/x) √(16x² - 7) dx is 4√(569) - 2.