Calculating Nitrogen Dioxide Levels in Long Beach on a May Day

[IHave]

Homework Statement

the amount of nitrogen dioxide, a brown gas that impairs breathing, present in the atmosphere on a certain day in may in the city of long beach is appoximated by:

A(t) = ((544) / (4 + (t - 4.5)^2) + 28 t is on interval [0, 11]

where A(t) is measured in pollutant standard index (PSI) and t is measured in hours with t = 0 corresponding to 7 A.M. What is the average amount of the pollutant present in the atmosphere between 7 A.M. and 2 P.M. on that day in the city?

Homework Equations

I know that this problem can be solved using trig substitution since that is the section in my book, in which it came out from.

The Attempt at a Solution

I've made several attempts to solve this problem through integration/trig sub. but i keep getting the incorrect value.

 

[Harrisonized]

Shift the interval.

∫_a^b f '(x-c) dx

= f(x-c)|_a^b

= f(b-c) -f(a-c)

= ∫_a-c^b-c f '(x) dx

After shifting, the trig-sub should be easy to detect.

 

[HallsofIvy]

I presume you know that the "average value" of an integrable function, f, between x= a and x= b, is
$$\frac{\int_a^b f(x)dx}{b- a}$$

The "shifting" that Harrisonized refers to is equivalent to the simple substitution u= x- 4.5. Once you have done that your function will involve $1/(4+ u^2$ and you should be able to recognize that immediately.

 

[IHave]

I presume you know that the "average value" of an integrable function, f, between x= a and x= b, is
$$\frac{\int_a^b f(x)dx}{b- a}$$

The "shifting" that Harrisonized refers to is equivalent to the simple substitution u= x- 4.5. Once you have done that your function will involve $1/(4+ u^2$ and you should be able to recognize that immediately.

thank you so much, i knew i was missing something 'cause the integral itself doesn't find the average. so i was missing the b-a component. thanks i was able to figure it out.

 

[IHave]

108 PSI if anyone was wondering.