# Math Help - Mean Value Theorem for integrals w/ log

1. ## Mean Value Theorem for integrals w/ log

An arched shaped monument is 570 ft high and has a 600 ft base. It can be modeled by the function $y=1140-285(e^{0.00439x}+e^{-0.00439x})$ where the base of the arch is[-300,300] and x and y are measured in feet. Find the average height of the arch above the ground.

I was able to do this but it took a while. What is the best way to to do this so I dont have to enter too many buttons on my calculator which personally increases my rate of error.

My questions: 1. Can I use ln on the integrand and what would it look like? 2. How do I deal with symmety for a LN()?

$y=1140-285(e^{0.00439x}+e^{-0.00439x})$

$avg value =\frac{1}{600}\int_{-300}^{300}(1140-285(e^{0.00439x}+e^{-0.00439x})) dx$

factor out 285 and distribute the minus sign

$avg value =\frac{1}{600}*285\int_{-300}^{300}(4-e^{0.00439x}-e^{-0.00439x}) dx$

Can I use a Natral Log? and does it look like this???

$LN(avg value) = LN[\frac{285}{600}]\int_{-300}^{300}(ln4-0.00439x+0.00439x)dx$

Specifically, I want to know do I use LN on the factored out numbers and the avgvalue?

If all is well then My next step would be to split up the intrgrand and check for symmetry. If I didnt previously apply the Ln, then all of the integrand was an even functon. It appers that $e^x$ is an even function but what about LN(x)

ln(-x)= no bueno

2. ## Re: Mean Value Theorem for integrals w/ log

Originally Posted by delgeezee
An arched shaped monument is 570 ft high and has a 600 ft base. It can be modeled by the function $y=1140-285(e^{0.00439x}+e^{-0.00439x})$ where the base of the arch is[-300,300] and x and y are measured in feet. Find the average height of the arch above the ground.

I was able to do this but it took a while. What is the best way to to do this so I dont have to enter too many buttons on my calculator which personally increases my rate of error.

My questions: 1. Can I use ln on the integrand and what would it look like? 2. How do I deal with symmety for a LN()?

$y=1140-285(e^{0.00439x}+e^{-0.00439x})$

$avg value =\frac{1}{600}\int_{-300}^{300}(1140-285(e^{0.00439x}+e^{-0.00439x})) dx$

factor out 285 and distribute the minus sign

$avg value =\frac{1}{600}*285\int_{-300}^{300}(4-e^{0.00439x}-e^{-0.00439x}) dx$

Can I use a Natral Log? and does it look like this???

$LN(avg value) = LN[\frac{285}{600}]\int_{-300}^{300}(ln4-0.00439x+0.00439x)dx$

Specifically, I want to know do I use LN on the factored out numbers and the avgvalue?

If all is well then My next step would be to split up the intrgrand and check for symmetry. If I didnt previously apply the Ln, then all of the integrand was an even functon. It appers that $e^x$ is an even function but what about LN(x)

ln(-x)= no bueno
note that the antiderivative of $e^{kx}$ is $\frac{1}{k} e^{kx}$ ... it does not involve a log

let $a = 0.00439$

$h_{avg} =\frac{285}{600} \int_{-300}^{300}(4-e^{ax}-e^{-ax}) dx$

$h_{avg} =\frac{285}{600} \left[4x- \frac{1}{a}e^{ax} + \frac{1}{a}e^{-ax} \right]_{-300}^{300}$

also note that you may take advantage of symmetry ...

$h_{avg} =\frac{570}{600} \left[4x- \frac{1}{a}e^{ax} + \frac{1}{a}e^{-ax} \right]_0^{300}$

3. ## Re: Mean Value Theorem for integrals w/ log

Originally Posted by delgeezee
An arched shaped monument is 570 ft high and has a 600 ft base. It can be modeled by the function $y=1140-285(e^{0.00439x}+e^{-0.00439x})$ where the base of the arch is[-300,300] and x and y are measured in feet. Find the average height of the arch above the ground.

I was able to do this but it took a while. What is the best way to to do this so I dont have to enter too many buttons on my calculator which personally increases my rate of error.

My questions: 1. Can I use ln on the integrand and what would it look like? 2. How do I deal with symmety for a LN()?

$y=1140-285(e^{0.00439x}+e^{-0.00439x})$

$avg value =\frac{1}{600}\int_{-300}^{300}(1140-285(e^{0.00439x}+e^{-0.00439x})) dx$

factor out 285 and distribute the minus sign

$avg value =\frac{1}{600}*285\int_{-300}^{300}(4-e^{0.00439x}-e^{-0.00439x}) dx$

Can I use a Natral Log? and does it look like this???

$LN(avg value) = LN[\frac{285}{600}]\int_{-300}^{300}(ln4-0.00439x+0.00439x)dx$
No. ln(a+ b) is NOT equal to ln(a)+ ln(b).

Specifically, I want to know do I use LN on the factored out numbers and the avgvalue?
I don't see why you would want to. The logarithm is not relevant. You want to integrate
$\frac{57}{120}\left(\int_{-300}^{300}4 dx- \int_{-300}^{300}e^{0.00439x}dx- \int_{-300}^{300}e^{-0.00439x}dx$
all of which are very easy integrals.

If all is well then My next step would be to split up the intrgrand and check for symmetry. If I didnt previously apply the Ln, then all of the integrand was an even functon. It appers that $e^x$ is an even function but what about LN(x)

ln(-x)= no bueno