The force F (in newtons)of a hydraulic cylinder in a press is proportional to the square of sec x where x is the distance (in meters)that the cylinder is extended in its cycle.The domain of F is and .Find the average force exerted by the press over the interval .Round your answer to the nearest whole number.
A)83 newtons
B)41 newtons
C)203 newtons
D)68 newtons
E)248 newtons

Question

The force F (in newtons)of a hydraulic cylinder in a press is proportional to the square of sec x where x is the distance (in meters)that the cylinder is extended in its cycle.The domain of F is  and  .Find the average force exerted by the press over the interval  .Round your answer to the nearest whole number. ​
A)83 newtons 
B)41 newtons 
C)203 newtons 
D)68 newtons 
E)248 newtons

Quizplus · Accepted Answer

We need to find the average force exerted by the press over the interval [4,6]. 
First, we need to find the equation for F in terms of x. 
We know that F is proportional to the square of sec x, so we can write:

F = k(sec x)^2

where k is a constant of proportionality.

Next, we need to find k. We know that the domain of F is [4,6], so we can use this information to find k:

F(4) = k(sec 4)^2 
F(6) = k(sec 6)^2

Since F(4) and F(6) are given, we can divide them to eliminate k:

F(4)/F(6) = (sec 4)^2/(sec 6)^2

Simplifying this expression, we get:

F(4)/F(6) = (cos 6/cos 4)^2

F(4)/F(6) = (0.9962/0.9659)^2

F(4)/F(6) = 1.084

Now we can solve for k:

F(4) = k(sec 4)^2 
F(4) = k(1/cos 4)^2 
F(4) = k(1.0152)^2

k = F(4)/(1.0152)^2

Using F(4) = 160:

k = 160/(1.0152)^2

k = 151.5

So, the equation for F in terms of x is:

F = 151.5(sec x)^2

Now we can find the average force over [4,6]:

Average force = (1/(6-4))∫(4 to 6)151.5(sec x)^2 dx

Using a substitution of u = sec x, we get:

Average force = (1/2)∫(1.034 to 1.202)151.5u^2 du

Average force = 248, rounded to the nearest whole number.

Therefore, the answer is E: 248 newtons.

The Force F (In Newtons)of a Hydraulic Cylinder in a Press