We first must establish the following relationship

$\displaystyle p''(t)=a(t)$ and $\displaystyle p'(t)=v(t)$

We now may note that 

$\displaystyle \int a(t)dt=v(t)+C$ or $\displaystyle \int4dt=4t+C=v(t)$

Since $\displaystyle v(t)=3t+C$

We must plug in our initial condition $\displaystyle v(0)=4$

Therefore our new velocity equation is $\displaystyle v(t)=4t+4$

We now may similarly integrate the velocity equation to find position.

$\displaystyle \int(4t+4)dt=2t^2+4t+C$

Plugging in our second initial condition $\displaystyle p(0)=-4$

We find our final equation to be:

$\displaystyle p(t)=2t^2+4t-4$

To find the velocity function, we integrate the acceleration function (the acceleration is the antiderivative of the velocity):

$\displaystyle v(t)=\int a(t)dt=\int [3t +\cos(t)]dt =\frac{ 3t^2}{2}+\sin(t)+C$

The rules of integration used were

$\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1}+C$, $\displaystyle \int \cos(t)dt= \sin(t)+C$

To solve for the integration constant, we plug in the given initial condition:

$\displaystyle v(0)=3=0+0+C$

$\displaystyle 3=C$

Our final answer is

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

To find the position function, we must integrate the velocity function, as velocity is the antiderivative of position:

$\displaystyle s(t)=\int v(t) dt = \int (10t+1)dt = 5t^2+t+C$

The following rule of integration was used:

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

Finally, we use the initial condition to solve for the integration constant:

$\displaystyle s(0)=0=0+0+C$

$\displaystyle C=0$

Our final answer is

$\displaystyle s(t)=5t^2+t$

In this problem, letting $\displaystyle s(t)$ denote the position of the particle and $\displaystyle v(t)$ denote the velocity, we know that $\displaystyle s''(t) = v'(t) = a(t)$. Integrating and working backwards we have,

$\displaystyle v(t) = \int a(t)dt = \int (-24t^2 + 6t - 8)dt = -8t^3 + 3t^2 -8t + C$

Plugging in our initial condition, $\displaystyle v(0) = 32$, we see immediately that $\displaystyle C = 32$.

Repeating the process again for $\displaystyle s(t)$, we find that 

 $\displaystyle s(t) = \int v(t)dt = \int (-8t^3 + 3t^2 -8t +32)dt = -2t^4 + t^3 - 4t^2 + 32t + C$

Plugging in our initial condition, $\displaystyle s(0) = 0$ (we started at the origin) we see that $\displaystyle C = 0$. This gives us a final equation

$\displaystyle s(t) = -2t^4 + t^3 - 4t^2 + 32t$. The problem asks for $\displaystyle s(1)$ which is simply $\displaystyle s(t) = -2(1)^4 + (1)^3 - 4(1)^2 + 32(1) = -4 + 1 - 4 + 32 = 27$

Find the integral which satisfies the specific conditions of $\displaystyle (\frac{\pi}{2}, 4)$

$\displaystyle \int4sin(t)-12cos(t)+3t^2dt$

To do this problem, we need to recall that integrals are also called anti-derivatives. This means that we can calculate integrals by reversing our integration rules.

Furthermore, to find the specific answer using initial conditions, we need to find our "c" at the end.

Thus, we can have the following rules.

$\displaystyle \int sin(t)dt = -cos(t)+c$

$\displaystyle \int cos(t)dt = sin(t)+c$

$\displaystyle \int t^ndt=\frac{t^{n+1}}{n+1}+c$

Using these rules, we can find our answer:

$\displaystyle \int4sin(t)-12cos(t)+3t^2dt$ 

Will become:

$\displaystyle -4cos(t)-12sin(t)+\frac{3t^3}{3}+c=-4cos(t)-12sin(t)+t^3+c$

And so our anti-derivative is:

$\displaystyle -4cos(t)-12sin(t)+t^3+c$

Now, let's find c. First set our above expression equal to y

$\displaystyle y= -4cos(t)-12sin(t)+t^3+c$

Next, plug in   for y and t. Then solve for c

$\displaystyle 4= -4cos(\frac{\pi}{2})-12sin(\frac{\pi}{2})+(\frac{\pi}{2})^3+c$

Looks a bit messy, but we can clean it up to get:

$\displaystyle 4= 0-12+3.88+c\rightarrow 4=-8.12+c\rightarrow c=12.12$

Now, to solve, simply replace c with 12.12

$\displaystyle y=-4cos(t)-12sin(t)+t^3+12.12$

The following equation is used for finding the Average Value of a Function:  $\displaystyle f_{avg}=\frac{1}{b-a} \int_{a}^{b} f(x) dx$. A rearrangement of this equation could be multiplying $\displaystyle (b-a)$ to both sides. Making this rearrangement, and substituting $\displaystyle f_{avg}=f(c)$ with $\displaystyle a< c< b$, results in the following: $\displaystyle \int_{a}^{b}f(x) dx=f(c)(b-a)$. Assuming $\displaystyle f(x)$ is continuous, this is the correct equation for the Mean Value Theorem for Integrals.

When finding the average value of a function, it is useful to keep the following formula in mind: $\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$. This equation is very helpful because it provides a simple way to determine the average value by substituting in values of the bounds and the function itself:

$\displaystyle f_{avg}=\frac{1}{3-1}\int_{1}^{3}2x^7+5x^2+4 dx=(12)[\frac{1}{4}x^8+\frac{5}{3}x^3+4x] |_1^3$$\displaystyle =(\frac{1}{2})(\frac{6561}{4}+45+12-1\frac{}{4}-\frac{5}{3}-4)=845.67$

When finding the average value of a function, it is useful to keep the following formula in mind: $\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$. This equation is very helpful because it provides a simple way to determine the average value by substituting in values of the bounds and the function itself:

$\displaystyle f_{avg}=\frac{1}{5-0}\int_{0}^{5}4e^{2x} dx=(\frac{1}{5})[4e^{2x} * \frac{1}{2}] |_0^5 =\frac{2}{5}e^{2(5)}-\frac{2}{5}e^{2(0)}=\frac{2}{5}(e^{10}-1)$

When finding the average value of a function, it is useful to keep the following formula in mind: $\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$. This equation allows the substitution of the function and interval to solve for the average value. While the function in this problem contains a trigonometric function, the same approach can be applied. Remember that the function is in terms of t, so the definite integral expression should likewise be in terms of $\displaystyle t$.

$\displaystyle f_{avg}=\frac{1}{\frac{\pi}{4}-0}\int_{0}^{\frac{\pi}{4}}4+2cos(2t) dt=(\frac{4}{\pi})[4t+sin(2t)] |_0^{\frac{\pi}{4}}$$\displaystyle =(\frac{4}{\pi})[\pi +1-0-0]=4+\frac{4}{\pi}$

When finding the average value of a function, it is useful to keep the following formula in mind: $\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx$. This equation allows the substitution of the function and interval to solve for the average value. While the function in this problem contains a trigonometric function, the same approach can be applied. Remember that the function is in terms of $\displaystyle t$, so the definite integral expression should likewise be in terms of $\displaystyle t$.

$\displaystyle f_{avg}=\frac{1}{\pi-0}\int_{0}^{\pi}6-(3t)sin(t^2) dt =\frac{1}{\pi}\int_{0}^{\pi}6-3tsin(t^2) dt$