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

$\displaystyle L=\int ds$

where ds is given by the equation below:

 $\displaystyle ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

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:

$\displaystyle \frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}$

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

$\displaystyle ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}$

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:

$\displaystyle L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}$

$\displaystyle =\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

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:

$\displaystyle \sqrt{x}=x^3\rightarrow x^3-\sqrt{x}=0\rightarrow x^\frac{1}{2}(x^\frac{5}{2}-1)=0$

$\displaystyle x^\frac{1}{2}=0\rightarrow x=0$

$\displaystyle x^\frac{5}{2}-1=0\rightarrow x^\frac{5}{2}=1\rightarrow x=1$

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:

$\displaystyle \bar{x}=\frac{1}{A}\int_{a}^{b}x(f(x)-g(x))dx$

$\displaystyle \bar{y}=\frac{1}{A}\int_{a}^{b}\frac{1}{2}([f(x)]^2-[g(x)]^2)dx$

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:

$\displaystyle A=\int_{a}^{b}(f(x)-g(x))dx=\int_{0}^{1}(\sqrt{x}-x^3)dx=[\frac{2}{3}x^\frac{3}{2}-\frac{1}{4}x^4]_{0}^{1}$

$\displaystyle =(\frac{2}{3}-\frac{1}{4})-(0)=\frac{8}{12}-\frac{3}{12}=\frac{5}{12}$

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:

$\displaystyle \bar{x}=\frac{1}{\frac{5}{12}}\int_{0}^{1}x(\sqrt{x}-x^3)dx=\frac{12}{5}\int_{0}^{1}(x^\frac{3}{2}-x^4)dx$

$\displaystyle =\frac{12}{5}[\frac{2}{5}x^\frac{5}{2}-\frac{1}{5}x^5]_{0}^{1}=\frac{12}{5}(\frac{1}{5})=\frac{12}{25}$

$\displaystyle \bar{y}=\frac{1}{\frac{5}{12}}\int_{0}^{1}\frac{1}{2}([\sqrt{x}]^2-[x^3]^2)dx=\frac{12}{5}\int_{0}^{1}(\frac{1}{2}x-\frac{1}{2}x^6)dx$

 $\displaystyle =\frac{12}{5}[\frac{1}{4}x^2-\frac{1}{14}x^7]_{0}^{1}=\frac{12}{5}(\frac{7}{28}-\frac{2}{28})=\frac{12}{5}(\frac{5}{28})=\frac{60}{140}=\frac{3}{7}$

So by evaluating our integrals, we can see that the center of mass of the region bounded by our two functions is   $\displaystyle \tiny (\frac{12}{25},\frac{3}{7})$.

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:

$\displaystyle S=\int (2\pi y)ds$

where ds is given by the following equation:

$\displaystyle ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

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:

$\displaystyle \frac{dy}{dx}=\frac{1}{2}(16-x^{2})^{-\frac{1}{2}}(-2x)=-\frac{x}{(16-x^{2})^{\frac{1}{2}}}$

$\displaystyle ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx=\sqrt{1+(-\frac{x}{\sqrt{16-x^{2}}})^{2}}dx=\sqrt{1+\frac{x^{2}}{16-x^{2}}dx$

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:

$\displaystyle ds=\frac{4}{\sqrt{16-x^{2}}}dx$

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:

$\displaystyle S=\int (2\pi y)ds=\int_{-2}^{2}2\pi (\sqrt{16-x^{2}})(\frac{4}{\sqrt{16-x^{2}}})dx$

$\displaystyle =\int_{-2}^{2}(8\pi)dx=8\pi (2)-8\pi (-2)=32\pi$

$\displaystyle \small \small W=\int (Fdx)$, so $\displaystyle \small W=\int(3x^{2}+x)dx$.

Both the terms of the force are power terms in the form $\displaystyle \small ax^{n}$, which have the integral $\displaystyle \small \frac{a}{n+1}x^{n+1}$, so the integral of the force is $\displaystyle \small x^{3}+x^{2}/2+c$.

We know 

$\displaystyle \small W(0)=0\rightarrow 0^{3}+0^{2}/2+c=0 \therefore c=0$.

This means 

$\displaystyle \small W(x)=x^{3}+x^{2}/2$.

By the fundamental theorem of calculus, we know that $\displaystyle \int_{0}^{4} v(t)dt = s(4) - s(0)$

So, $\displaystyle s(4) = s(0) + \int_{0}^{4}v(t)dt$

$\displaystyle s(4) = 0 + \left [ \frac{5}{4}x^2 + 3x \right ]$ evaluated from $\displaystyle x = 0$ to $\displaystyle x=4$

$\displaystyle s(4) = (20 + 12) - (0 + 0) = 32 \ m$ 

The distance traveled by an object over an interval of time is the total area under the velocity curved during the interval. Therefore, we need to evaluate the definite integral

$\displaystyle \int_{0}^{4}2\sqrt{t}dt=\int_{0}^{4}2t^\frac{1}{2}dt$

Rewriting expressions written using radical notation using fractional exponents can make applying the power rule to find the antiderivative easier.

Evaluating this integral using the power rule,

$\displaystyle \int t^n=\frac{t^{n+1}}{n+1}+c$, we find:

$\displaystyle \int_{0}^{4}2t^\frac{1}{2}dt=\frac{2}{3}\cdot 2(4^\frac{3}{2}) -\frac{2}{3}\cdot 2(0^\frac{3}{2})$

$\displaystyle =\frac{4}{3}\cdot (8)=\frac{32}{3}\ meters$

The work done moving an object is the intergral of force applied as a function of distance,

 $\displaystyle W(x)=\int F(x)dx$.

To find the work done moving the object from x=1 to x=4, evaluate the definite interval using x=1 and x=4 as the limits of integration. Recall that the antiderivative of a polynomial is found using the power rule, 

$\displaystyle \int x^n=\frac{nx^{n+1}}{n+1}+c$. 

$\displaystyle \int_{1}^{4}(2x^2+3)dx=\frac{2(4^3)}{3}+3(4)-\frac{2(1^3)}{3}+3(1)$

$\displaystyle =\frac{164}{3}-\frac{11}{3}=\frac{153}{3}=51 joules$

The position function is the second antiderivative of acceleration. First, integrate the acceration function to find the velocity function. Since the acceleration function is a polynomial, the power rule is needed. Recall that the power rule is 

$\displaystyle \int{x^ndx}=\frac{x^{n+1}}{n+1}+c$.

$\displaystyle v(t)=\int{4t}dt=\frac{4t^2}{2}+c=2t^2+c$.

Next, we can find the the value of the constant $\displaystyle c$ using the fact that $\displaystyle v(0)=1$.

$\displaystyle v(0)=1=2(0^2)+c = c$, so $\displaystyle c = 1$. The velocity function is therefore $\displaystyle v(t)=2t^2+1$.

Now integrate the velocity function to find the position function.

$\displaystyle s(t)=\int{(2t^2+1)dt}=\frac{2t^3}{3}+t+c$.

Use the fact that $\displaystyle s(0)=0$ to find the constant $\displaystyle c$. 

$\displaystyle s(0)=0=\frac{2(0^3)}{3}+0+c=c,\rightarrow c=0$.

The velocity function is 

$\displaystyle s(t)=\frac{2t^3}{3}+t$

Integrating $\displaystyle r(t)$ over an interval $\displaystyle [a,b]$ will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral 

$\displaystyle \int_{0}^{5}2e^{0.5t}dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let$\displaystyle u=0.5t, du=0.5dt, \ and\ 2du=dt$. We should also express the limits of integration in terms of $\displaystyle u$. When $\displaystyle t = 0, u=0$, and when $\displaystyle t=5, u=2.5.$ Making these substitutions leads to the integral

$\displaystyle \int_{0}^{2.5}4e^udu$.

To evaluate this, you must know the antiderivative of an exponential function.

In general, 

$\displaystyle \int e^xdx=e^x+c$.

Therefore, 

$\displaystyle \int_{0}^{2.5}4e^udu=4e^u|_{0}^{2.5}=4e^{2.5}-4\approx 44.7$. 

This tells us that the temperature rose by approximately $\displaystyle 44.7$ degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at $\displaystyle t=5$ minutes is 

$\displaystyle 75+44.7=119.7$ degrees.

In order to find the velocity function, we must integrate the accleration function:

$\displaystyle \int (16t^2+48)dt=\frac{16}{3}t^3+48t+C$

We used the rule

$\displaystyle \int x^ndx=\frac{1}{n+1}x^{n+1}+C$ 

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that 

$\displaystyle v(0)=30$

so we plug in zero into the velocity function and solve for C:

$\displaystyle v(0)=30=\frac{16}{3}(0)^3+48(0)+C$

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

$\displaystyle v(t)=\frac{16}{3}t^3+48t+30$