How do you find the average value of #f(x)=x/sqrt(x^2+1), 0

Question

How do you find the average value of #f(x)=x/sqrt(x^2+1), 0<=x<=4#?

Samantha Adkison · Accepted Answer

The average value of #f(x)# on #0 ≤ x ≤ 4# is #1/4(sqrt(17) - 1)#, which is approximately #0.78#

The average value of a continuous function #f(x)# on #[a, b]#, is given by the formula #A = 1/(b - a) int_a^b f(x) dx#.

#A = 1/(4 - 0)int_0^4 x/sqrt(x^2 + 1)dx#

#A = 1/4int_0^4 x/sqrt(x^2 +1)dx#

This now becomes a trig substitution problem. Let #x = tantheta#. Then #dx = sec^2theta d theta#. Adjust the bounds of integration accordingly.

#A = 1/4int_0^(tan4) tantheta/sqrt(tan^2theta + 1) * sec^2theta d theta#

Use the pythagorean identity #tan^2alpha + 1 = sec^2alpha#.

#A = 1/4int_0^(tan4) tantheta/sqrt(sec^2theta) * sec^2theta d theta#

#A = 1/4int_0^(tan4) tantheta/sectheta * sec^2theta#

#A = 1/4int_0^(tan4) tanthetasectheta#

This is a known integral.

#A = 1/4[sectheta]_0^(tan4)#

We know from our initial substitution that #x/1 = tantheta#. This means that #sqrt(x^2 + 1)# is the hypotenuse. That's to say #sectheta = sqrt(x^2 + 1)#.

#A = 1/4[sqrt(x^2 +1)]_0^4#

Evaluate this using the 2nd fundamental theorem of calculus.

#A = 1/4(sqrt((4)^2 + 1) - sqrt(0^2 + 1))#

#A = 1/4sqrt(17) - 1/4#

#A = 1/4(sqrt(17) - 1)#

Hopefully this helps!

Evelyn Agard · Answer

Alternative to get #1/4(sqrt17-1)#:

Like HSBC244 said, The average value of a continuous function #f(x)# on #[a, b]#, is given by the formula #A = 1/(b - a) int_a^b f(x) dx#. #A = 1/(4 - 0)int_0^4 x/sqrt(x^2 + 1)dx# #A = 1/4int_0^4 x/sqrt(x^2 +1)dx# Instead of doing a trig substitution like HSBC244, you can do a u substitution instead. Set #u=x^2+1# which makes #du=2x#. Multiply the integral by #2/2# to get the 2 sufficient for the u substitution and put the 2 in the denominator on the outside. #A = 1/8int_1^17 1/sqrt(u)du# Note that the limits of integration changed because we were doing a u substitution. 0 became 1 because: #0^2+1=1# Similarly plugging in 4 became 17 because: #4^2+1=17# Now rewrite it so it's easier to integrate by treating the square root in the denominator: #A = 1/8int_1^17 u^(1/2)# This integral becomes #A = 1/8[2sqrt(u)]_1^17# Take the 2 out and plug in 17 and 1. As you can see, this will give you the same answer as HSBC244's answer but does not require a trig substitution.

Ezekiel Alexander · Answer

undefined To find the average value of $ f(x) = \frac{x}{\sqrt{x^2 + 1}} $ over the interval $ 0 \leq x \leq 4 $, follow these steps:

1. Compute the definite integral of $ f(x) $ over the given interval.
2. Divide the result from step 1 by the length of the interval.

The average value of $ f(x) $ over the interval $ 0 \leq x \leq 4 $ is approximately 1.3333.

How do you find the average value of #f(x)=x/sqrt(x^2+1), 0&lt;=x&lt;=4#?

How do you find the average value of #f(x)=x/sqrt(x^2+1), 0<=x<=4#?