This is a Second Fundamental Theorem of Calculus problem.  Since the derivative cancels out the integral, we just need to plug in the bounds into the function (top bound - bottom bottom) and multiply each by their derivative.

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Differentiate: 

Use the Second Fundamental Theorem of Calculus:

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If the function  is continuous on an open interval  and if  is in , then for a function  defined by , we have . 

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To understand how to apply the second FTOC, notice that  in our case is a function of a function. The chainrule will therefore have to be applied as well. To see how this works, let . The function can now be written in the form: 

Now we can go back to writing in terms of , the derivative is, 

We plug in  into our formula and determine the external force .

Taking the derivative of each side with respect to , and applying the fundamental theorem of calculus:

Solving for ,

In order to solve this, we substitute  into our equation and solve for :

Taking the derivative of each side with respect to , we get that:

Using the fundamental theorem of calculus:

Solving for :

Since ,

The Fundamental Theorem of Calculus tells us that 

therefore,

From here plug in 0 into this equation.

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

where ds is given by the equation below:

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

Now we can plug this into the given equation to find ds:

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

Before we can set up any sort of integral, we must find out where our two functions intersect, which tells of what the bounds of our region are. To do this, we set our functions equal to each other and solve for the x values at which they intersect:

Our next step is to use the formulas for the x and y coordinates of any region's center of mass, which are given below:

We know the bounds of our region, as well as the functions f(x) and g(x) that make up its upper and lower boundaries, respectively, so the only thing we have left to do is calculate the area of our region using the following integral:

Now that we know the area, bounds, and functions f(x) and g(x) that make up our region, we can simply plug these values into the formulas for the x and y coordinates of the center of mass for the region:

So by evaluating our integrals, we can see that the center of mass of the region bounded by our two functions is   .

Before we can begin the problem, we must remember the formula for finding the surface area of a solid formed from rotation about the x-axis:

where ds is given by the following equation:

So in order to set up our integral for surface area, we must first find ds, which requires us to evaluate the derivative of our function:

In order to add the two terms in our radicand, we must find their common denominator. Because one of our terms is just 1, we can simplify the equation by writing 1 as (16-x^2)/(16-x^2), which allows us to add the two terms together and gives us:

Now we have our expression for ds in terms of x, and we can write the y in the formula for surface area by replacing it with its equation in terms of x given in the problem statement:

, so .

Both the terms of the force are power terms in the form , which have the integral , so the integral of the force is .

We know 

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This means 

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By the fundamental theorem of calculus, we know that 

So, 

 evaluated from  to