Final answer:
To find the value of c for which the average value of the function v(x)=6/x² on the interval [1,c] is equal to 1, we set up an integral formula, solved the integral, got a quadratic equation, and found that c = 6 is the solution.
Step-by-step explanation:
The question asks to find the value of c for which the average value of the function v(x) = 6/x² on the interval [1,c] is equal to 1.
The average value of a function f(x) on the interval [a,b] is given by the integral formula:
\[\frac{1}{b-a}\int_{a}^{b} f(x) dx\]
Here, a = 1 and b = c. Plugging our function into this formula, we have:
\[1 = \frac{1}{c-1}\int_{1}^{c} \frac{6}{x²} dx\]
Integrating 6/x², we get:
\[1 = \frac{1}{c-1}[-6/x]_{1}^{c}\]
\[1 = \frac{1}{c-1}(-6/c + 6)\]
Multiplying both sides by c-1 and simplifying gives us:
\[c - 1 = 6 - \frac{6}{c}\]
\[c^2 - c = 6c - 6\]
\[c^2 - 7c + 6 = 0\]
This is a quadratic equation, which factors to:
\[(c - 6)(c - 1) = 0\]
So, c = 6 or c = 1. Since we already know that a = 1, c = 6 is our solution.