If your solution to a stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
010. `query 10
Question: `q5.5.1 (previously 5.5.23 (was 5.5.28)) area in region defined by y=8/x, y = x^2, x = 1, x = 4
Your solution:
Confidence Assessment:
Given Solution:
`a These graphs intersect when 8/x = x^2, which we solve to obtain x = 2.
For x < 2 we have 8/x > x^2; for x > 2 the inequality is the reverse.
So we integrate 8/x - x^2 from x = 1 to x = 2, and x^2 - 8 / x from x = 2 to x = 4.
Antiderivative are 8 ln x - x^3 / 3 and x^3 / 3 - 8 ln x. We obtain
8 ln 2 - 8/3 - (8 ln 1 - 1/3) = 8 ln 2 - 7/3 and
64/3 - 8 ln 4 - (8 ln 2 - 8/3) = 56/3 - 8 ln 2.
Adding the two results we obtain 49/3. **
STUDENT QUESTION
I integrated
(8 / x - x^2) from x = 1 to 4
I see what is going on, but how do I know when to use this exam and not what I used.
INSTRUCTOR RESPONSE
You should begin by constructing graphs of y = 1/x and y = x^2 on the interval 1 < x < 4. You should refer to these graphs as you work through the following:
To find the area between the graphs of two functions you always have to integrate the quantity (greater function - lesser function).
This means you have to determine which function is greater and which is lesser.
The answer to this question might the same for the entire interval, but it might also vary over different parts of the interval.
For the present functions, one is greater on part of the given interval, the other is greater on the other part of that interval.
As you see it is important to construct the graphs of the two functions to get an idea which is greater where.
You should actually construct the graphs so you will understand how the formulas for the functions determine how they behave.
The calculator would show you the graphs, but the calculator is not a legitimate way to justify your construction of a graph on a test or in homework.
The key idea
is as follows:
Assuming that f(x) and g(x) are continuous functions, then f(x) - g(x) is a
continuous function.
Where f(x) is greater, this function is positive.
Where g(x) is greater, this function is negative.
If the functions are reasonably well behaved (as they will be in this course), then f(x) - g(x) could be positive on some intervals and negative on others.
On intervals where f(x) - g(x) is positive, f(x) is the greater function and you will integrate f(x) - g(x).
On intervals where f(x) - g(x) is negative, g(x) is the greater function and you will integrate g(x) - f(x).
So if we can find the intervals where f(x) - g(x) is positive, and the intervals where negative, we can easily set up our integrals.
You should look at your graphs of f(x) and g(x), and use your graph to construct a graph of f(x) - g(x). You will see that your newly constructed graph is positive for part of the interval from x = 1 to x = 4, and negative for another part of this interval.
To find the
intervals, we observe that a continuous function can go from positive to
negative or from negative to positive only by taking the value 0 at the
point where it makes the change.
So if we can find the x values where f(x) - g(x) = 0, we are just about home
free.
In the present case, we can let f(x) = 8/x and g(x) = x^2.
One potential problem is that we are relying on the continuity of f(x) - g(x). However f(x) = 8 / x isn't always continuous; at x = 0 the function isn't defined, so it isn't continuous at x = 0. Fortunately the interval we are considering here is 1 < x < 4, which doesn't include x = 0. So on the interval we are considering, both functions are continuous.
Thus f(x) - g(x) = 8 / x - x^2 and our equation f(x) - g(x) = 0 becomes
8/x - x^2 = 0.We easily solve this equation (multiply both sides by x to get 8 - x^3 = 0 so x^3 = 8 and therefore x = 2).
Our solution is x = 2, so on the interval 1 < x < 4, the function 8/x - x^2 can change sign only at x = 2.
This splits our interval 1 < x < 4 into two subintervals, 1 < x < 2 and 2 < x < 4.
For 1 < x < 2 the function 8/x - x^2 has just one sign, and for 2 < x < 4 it just one sign. We can easily determine the signs on these intervals by substitution:
x = 3/2 lies in the interval 1 < x < 2, and when x = 3/2 our function 8/x - x^2 takes value 8/(3/2) - (3/2)^2 = 16/3 - 9/4 = (64 - 27) / 12, which is clearly positive.
So for 1 < x < 2, 8/x - x^2 is positive and 8/x must therefore be the greater function.
Thus the area on this interval is integral ( (9/x - x^2) , x from 1 to 2).
x = 3 lies in the interval 2 < x < 4, and when x = 3 our function 8/x - x^2 takes value 8/3 - 3^2 = 8/3 - 9 = (8 - 27) / 3, which is clearly negative.
So for 2 < x < 4 the function 8/x - x^2 is negative and x^2 is therefore the greater.
Thus the area on this interval is integral ( (x^2 - 8/x), x from 2 to 4).
Our total area is therefore
total area = integral ( (9/x - x^2) , x from 1 to 2) + integral ( (x^2 - 8/x), x from 2 to 4).
You should of course write out these integrals in standard form.
Notes:
At x = 2 our function 8/x - x^2 takes value 0. This means that the graphs of 8/x and x^2 must intersect at x = 2.
Another way to find the intersection point would just be to set the two functions equal. 8/x = x^2 when 8 = x^3, so they intersect when x = 2.
A simple graph of the functions won't necessarily show that the intersection point occurs when x = 2, but it will clearly show that 8/x is greater to the left of the intersection point, and x^2 is greater to the right.
This would tell you which function to subtract from which on which interval.
Self-critique (if necessary):
Self-critique Rating:
Question: `q5.5.4 (previously 5.5.44 (was 5.5.40) ) demand p1 = 1000-.4x^2, supply p2=42x
Your solution:
Confidence Assessment:
Given Solution:
`a 1000-.4x^2 = 42x is a quadratic equation. Rearrange to form
-.4 x^2 - 42 x + 1000 = 0 and use the quadratic formula.
You get x = 20
At x = 20 demand is 1000 - .4 * 20^2 = 840, supply is 42 * 20 = 840.
The demand and supply curves meet at (20, 840).
The area of the demand function above the equilibrium line y = 840 is the integral of 1000 - .4 x^2 - 840 = 160 - .4 x^2, from x = 0 to the equlibrium point at x = 20. This is the consumer surplus.
The area of the supply function below the equilibrium line is the integral from x = 0 to x = 20 of the function 840 - 42 x. This is the producer surplus.
The consumer surplus is therefore integral ( 160 - .4 x^2 , x from 0 to 20) = 2133.33 (antiderivative is 160 x - .4 / 3 * x^3). *&*&
Self-critique (if necessary):
Self-critique Rating: