Finding the average using integration

[chris_0101]

Homework Statement

Find c such that fave = f(c)
c1 (smaller value) = ?
c1 (smaller value) = ?

f(x) = 6x/(1 + x2)2

Average value (i.e. the integral) was found to be 12/5

Homework Equations

12/5 = 6x/(1 + x2)2

The Attempt at a Solution

12/5 = 6x/(1 + x²)²

12/5 = 6c/(1 + c²)²
12/5 = 6c/(1 +2c² + c⁴)
12(1 +2c² + c⁴) = 5(6c)
12 +24c² + 12c⁴ = 30c

12 - 30c +24c² + 12c⁴ = 0

Now, What I would do from this point on is to simply find a possible value for c, the proceed and use synthetic division to solve for the other values of c. However, this will produce 4 values for c instead of 2, which are required.

If anyone could shed some light on this question, that would greatly be appreciated

thanks

 

[gb7nash]

12 - 30c +24c² + 12c⁴ = 0

Does the problem specifically say there are only two distinct answers? If so, it is still possible to get just two distinct roots (they'll just be repeated roots). Assuming you did everything right up to this point, I would suggest trying the rational root theorem. If that fails, you could always try to quartic formula, but it's a crazy formula.

 

[Mark44]

Homework Statement

Find c such that fave = f(c)
c1 (smaller value) = ?
c1 (smaller value) = ?

f(x) = 6x/(1 + x2)2

Average value (i.e. the integral) was found to be 12/5

Homework Equations

12/5 = 6x/(1 + x2)2

The Attempt at a Solution

12/5 = 6x/(1 + x²)²

12/5 = 6c/(1 + c²)²
12/5 = 6c/(1 +2c² + c⁴)
12(1 +2c² + c⁴) = 5(6c)
12 +24c² + 12c⁴ = 30c

12 - 30c +24c² + 12c⁴ = 0

Now, What I would do from this point on is to simply find a possible value for c, the proceed and use synthetic division to solve for the other values of c. However, this will produce 4 values for c instead of 2, which are required.

If anyone could shed some light on this question, that would greatly be appreciated

thanks

When you found the average value, you evaluated a definite integral, with limits of integration. Of the four possible values from your quartic equation, pick the one that is in the same interval as you integrated on.