calculus problem?

Question

Find the average value of the following function on the given interval: f(x)= x/(sqrt(x^2+16)), on interval [0,3]. I need to calculate the average value of f on [0,3]

Can anyone explain how to work this problem???

thanks for the help

Answer 1

This problem is quite amusing, as knowledge of the most well-known Pythagorean triple --- 3,4,5 --- enables the final result to be evaluated easily.

You CAN NOT simply work out the average value of a NON-LINEAR FUNCTION by just averaging its end values (Tut, tut, gopal and Aegor R)! You have to evaluate the function's integral, and then divide by the length of the range (in this case, 3).

So, the indefinite integral of this nicely chosen function is:

(x^2 + 16)^(1/2) + C; NOTE that this is the same as
(x^2 + 4^2)^(1/2) + C (Hint, hint!)

When working out the value, the C's will cancel out.

For x = 3, the value is obviously 5 (Pythagorean triple).

For x = 0, the value is 4.

So, the integral is just 5 - 4 = 1 (It's also 1!; a factorial joke)

Thus, the average value is 1/3 or 0.33333... for those fractionally challenged.

CHECK: To check on the reasonableness of this result, note that the form of the function near x = 0 is the straight line y = x/4. Were this maintained throughout the full range, the mean would be 3/8 or 0.375. But the function slowly curves down from this straight line. At x = 3, its value is 3/5, which would correspond to an intersection with the analogous straight line through the origin, y = x/5; the mean value for this latter function would be 3/10 or 0.300.

The original function always lies between these two comparison straight lines in the given range; so the mean value of the original function (0.333...) clearly must lie between 0.300 and 0.375 --- and it does. BINGO!

Live long and prosper.

Answer 2

Avg = [1/(3 - 0)]Integral{x/(sqrt(x^2 + 16))}

Evaluate this...that is the average of f(x) over [0,3].


==> Avg = (1/3)*Sqrt(x^2 + 16)

==> evaluate over the interval [0,3]........(1/3)[Sqrt(16) - Sqrt(9 + 16)]

==> (1/3)[9 - 5]

==> 4/3

(done)

This makes sense because if you look at the plot, it increases steadily from zero to 3 (the endpoints of the curve). Other answers claim the average is 0.2.....how can that possibly be? The average of a functrion must take into account the fact that the curve has greater contribution as the value of 'x' increases. An average of 1/5 gives 0.2, that is almost at the beginning of the plot. That is not correct.

Think about this. If you have 5 numbers: 1, 2, 3, 4, 5, the avg is 3 right? What happens if I increase these as I go?

1, 3, 11, 37, 81

the average is now shifted HIGHER (26.6). This is why (1/5) can't be the average to your function. Does this make sense?

Answer 3

You have to find the value at 0, and the value at 3. That will give you the change in y, or the change in f(x). Then you divide that by the change in x, which is 3 (Just 3-0)

f(0) =0/sq(0+16) = 0/sq(16)= 0/4 = 0
f(x) =3/sq(16+3^2) = 3/sq(16+9) = 3/(sq(25) = 3/5

Divide this change, 3/5 by three and get 1/5

1/5 is your answer.

Hope this helps.

Answer 4

f (x) = x^3 - 12x + a million . . . the 1st by-product set to 0 shows turning or table certain factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the 2d by-product evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== valuable value shows x=2 is a close-by minimum f ' ' (-2) = 6*(-2) = -12 <== unfavorable value shows x=-2 is a close-by optimal a.) x = - 2 is a optimal, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is reducing x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the 2d by-product set to 0 shows inflection factors, or the place concavity changes 6x = 0 x = 0 <=== inflection factor x = - 2 is a optimal, so would desire to be concave down concavity changes on the inflection factor(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up

Answer 5

f(0)=0/sqrt((0)^2 + 16)
=0

f(3)=3/sqrt((3)^2 + 16)
=3/sqrt(9 +16)
=3/sqrt25
=3/5

(3/5)/3=3/15=1/5

Answer 6

f(0)=0
f(3)=3/5
average value=(3/5)/3
=1/5