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

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

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

factor out 285 and distribute the minus sign

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

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

$\displaystyle 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 $\displaystyle e^x$ is an even function but what about LN(x)

ln(-x)= no bueno

Re: Mean Value Theorem for integrals w/ log

Quote:

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

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

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

factor out 285 and distribute the minus sign

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

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

$\displaystyle 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 $\displaystyle e^x$ is an even function but what about LN(x)

ln(-x)= no bueno

note that the antiderivative of $\displaystyle e^{kx}$ is $\displaystyle \frac{1}{k} e^{kx}$ ... it does not involve a log

let $\displaystyle a = 0.00439$

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

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

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

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

Re: Mean Value Theorem for integrals w/ log

Quote:

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

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

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

factor out 285 and distribute the minus sign

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

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

$\displaystyle 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).

Quote:

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

$\displaystyle \frac{57}{120}\left(\int_{-300}^{300}4 dx- \int_{-300}^{300}e^{0.00439x}dx- \int_{-300}^{300}e^{-0.00439x}dx$

all of which are very easy integrals.

Quote:

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 $\displaystyle e^x$ is an even function but what about LN(x)

ln(-x)= no bueno