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:

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

The force of gravity is proportional to the mass of the object and acceleration of the object. 

$\displaystyle F=ma= (10*-10)\frac{kg*m}{s^2}= -100N$

Since the block fell down 5 meters, its final position is $\displaystyle -5$ and initial position is $\displaystyle 0$.

$\displaystyle \\ W=\int_{0}^{-5}-100dx\\ \\=\frac{-100x}{2}|_{0}^{-5} \\ \\=-50x|_{0}^{-5}\\ \\=-50(-5)--50(0)\\ \\=250J$

Evaluate the following integral and find the specific function which satisfies the given initial conditions:

$\displaystyle f(x)=\int3x^2-e^x+2dx$

$\displaystyle F(0)=3$

To solve this problem,we need to evaluate the given integral, then solve for our constant of integration.

Let's begin by recalling the following integration rules:

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

$\displaystyle 2) \int e^xdx=e^x+c$

Using these two, we can integrate f(x)

$\displaystyle f(x)=\int3x^2-e^x+2dx$

$\displaystyle \int3x^2-e^x+2dx=\frac{3x^3}{3}-e^x+2x+c=x^3-e^x+2x+c$

SO, we get:

$\displaystyle F(x)=x^3-e^x+2x+c$

We are almost there, but we need to find c. To do so, plug in our initial conditions and solve:

$\displaystyle 3=0^3-e^0+2(0)+c$

$\displaystyle 3=-1+c$

$\displaystyle c=4$

So, our answer is:

$\displaystyle F(x)=x^3-e^x+2x+4$

From calculus, we know the volume of an irregular solid can be determined by evaluating the following integral:

$\displaystyle V=\int_{a}^{b}A(x)dx$

Where A(x) is an equation for the cross-sectional area of the solid at any point x. We know our bounds for the integral are x=1 and x=4, as given in the problem, so now all we need is to find the expression A(x) for the area of our solid.

From the given bounds, we know our unrotated region is bounded by the x-axis (y=0) at the bottom, and by the line y=x^2-4x+5 at the top. Because we are rotating about the x-axis, we know that the radius of our solid at any point x is just the distance y=x^2-4x+5. Now that we have a function that describes the radius of the solid at any point x, we can plug the function into the formula for the area of a circle to give us an expression for the cross-sectional area of our solid at any point:

$\displaystyle A(x)=\pi r^2=\pi (x^2-4x+5)^2=\pi (x^4-8x^3+26x^2-40x+25)$

We now have our equation for the cross-sectional area of the solid, which we can integrate from x=1 to x=4 to find its volume:

$\displaystyle V=\int_{1}^{4}\pi (x^4-8x^3+26x^2-40x+25)dx$

$\displaystyle =\pi [\frac{1}{5}x^5-2x^4+\frac{26}{3}x^3-20x^2+25x]_{1}^{4}=\frac{78\pi }{5}$

Write the formula for cylindrical shells, where $\displaystyle r$ is the shell radius and $\displaystyle h$ is the shell height.

$\displaystyle V=\int_{a}^{b} 2\pi rh \:dy$

Determine the shell radius.

$\displaystyle r=y$

Determine the shell height.  This is done by subtracting the right curve, $\displaystyle x=4$, with the left curve, $\displaystyle x=y^2$.

$\displaystyle h=4-x=4-y^2$

Find the intersection of $\displaystyle x=y^2$ and $\displaystyle x=4$ to determine the y-bounds of the integral.

$\displaystyle 4=y^2$

$\displaystyle 2=y$

The bounds will be from 0 to 2.  Substitute all the givens into the formula and evaluate the integral.

$\displaystyle V=\int_{a}^{b} 2\pi rh \:dy = \int_{0}^{2} 2\pi \cdot y\cdot \left ( 4-y^2)dy$

$\displaystyle V= 2\pi \int_{0}^{2} \left ( 4y-y^3\right ) dy$

$\displaystyle V=2\pi \left[ 2y^2 - \frac{1}{4}y^4\right]_{0}^{2} = 2\pi \left[2(2)^2 -\frac{1}{4}(2)^4\right] = 2\pi[8-4]= 8\pi$

Since we are revolving a region of interest around a horizontal line $\displaystyle {}y = 0$, we need to express the inner and outer radii in terms of x.

Recall the formula:

$\displaystyle Volume = \pi \int_{b}^{a} [(\textup{outer radius})^{2} - (\textup{inner radius})^{2}]dx$

The outer radius is $\displaystyle y = sin(x)$ and the inner radius is $\displaystyle y = 0$. The x-limits of the region are between $\displaystyle x=0$ and $\displaystyle x=\pi$. So the volume set-up is:

$\displaystyle V = \pi \int_{0}^{\pi} sin^{2}x$

Using trigonometric identities, we know that:    

$\displaystyle sin^{2}x = \frac{1}{2} - \frac{1}{2}cos 2x$

Hence:

$\displaystyle V = \pi \int_{0}^{\pi} sin^2x dx =\pi \int_{0}^{\pi} \left(\frac{1}{2} - \frac{1}{2}cos2x\right) dx$

$\displaystyle = \pi \left[\frac{1}{2}x - \frac{1}{4}sin2x\right]_{0}^{\pi} = \frac{1}{2}\pi^2$

The rate at which the height changes is $\displaystyle \small \frac{\mathrm{d} h}{\mathrm{d} t}$, which means $\displaystyle \small \small \frac{\mathrm{d} h}{\mathrm{d} t}=5t^{1/2}$.

To find the height after nine seconds, we need to integrate to get $\displaystyle {}\small h(t)$.

We can multiply both sides by $\displaystyle \small dt$ to get $\displaystyle \small dh=5t^{1/2}dt$ and then integrate both sides.

This gives us 

$\displaystyle \small h=\frac{10t^{3/2}}{3}+c$.

Since the cup is empty at $\displaystyle \small t=0, h(0)=0$, so $\displaystyle \small c=0$.

This means $\displaystyle \small \small h(9)=10(9)^{3/2}/3=90$. No units were given in the problem, so leaving the answer unitless is acceptable.

Write the washer's method.

$\displaystyle V=\int_{a}^{b}\pi[R(y)^2-r(y)^2]\:dy$

Set the equations equal to each other to determine the bounds.

$\displaystyle 9x=x^3$

$\displaystyle 9=x^2$

$\displaystyle x=3$

The bounds are from 0 to 3.

Determine the big and small radius.  Rewrite the equations so that they are in terms of y.

$\displaystyle R(y)= \sqrt[3]{y}$

$\displaystyle r(y)=\frac{y}{9}$

Set up the integral and solve for the volume.

$\displaystyle V=\int_{0}^{3}\pi[(\sqrt[3]{y})^2-(\frac{y}{9})^2]\:dy$

$\displaystyle V=\int_{0}^{3}\pi[y^{\frac{2}{3}}-\frac{y^2}{81}]\:dy=\pi[\frac{3}{5}y^\frac{5}{3}-\frac{y^3}{243}]_{0}^{3}=11.41353$

The volume to the nearest integer is:  $\displaystyle 11$

Write the volume formula for cylindrical shells.

$\displaystyle V=\int_{a}^{b}2\pi (\textup{shell radius})(\textup{shell height}) dy$

The shell radius is $\displaystyle y$.

The shell height is the function in terms of $\displaystyle y$.  Rewrite that equation.

$\displaystyle y=\sqrt[3]{x}$

$\displaystyle x=y^3$

$\displaystyle \textup{shell height}: 1-y^3$

The bounds lie on the y-axis since the thickness variable is $\displaystyle dy$.  This is from 0 to 1, since the intersection of the line $\displaystyle x=1$ and $\displaystyle y=\sqrt[3]{x}$ is at $\displaystyle (1,1)$.  

Substitute all the values and solve for the volume.

$\displaystyle V=\int_{0}^{1}2\pi (y)(1-y^3) dy=\int_{0}^{1}2\pi (y-y^4)dy$

$\displaystyle V=2\pi \int_{0}^{1}(y-y^4)dy= 2\pi \left[\frac{y^2}{2}-\frac{y^5}{5}\right]_{0}^{1}= 2\pi\left[\frac{1}{2}-\frac{1}{5}\right]= \frac{3}{5}\pi$