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

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

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=5\\&f(1)=-19\\&f(4)=5\\&f(7)=-6\\&f(10)=-18\\&f(13)=0\\&\\&\text{Although the total interval has a length of }15\\&\text{Notice the points are equidistant. This distance is our subinterval }3\\&\text{and since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{13}f(x)\approx3[(-19)+(5)+(-6)+(-18)+(0)]\\&\int_{-2}^{13}f(x)\approx-114\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=-13\\&f(-5)=-19\\&f(-2)=6\\&f(1)=-9\\&\text{Although the total interval has a length of }9\\&\text{Notice the points are equidistant. This distance is our subinterval: }3\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-8}^{1}f(x)\approx3[(-19)+(6)+(-9)]\\&\int_{-8}^{1}f(x)\approx-66\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=-2\\&f(3)=-15\\&f(8)=14\\&f(13)=15\\&f(18)=-9\\&f(23)=-12\\&\text{Although the total interval has a length of }25\\&\text{Notice the points are equidistant, with }5\text{ intervals between. Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{23}f(x)\approx5[(-15)+(14)+(15)+(-9)+(-12)]\\&\int_{-2}^{23}f(x)\approx-35\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-7)=18\\&f(-2)=-17\\&f(3)=-16\\&f(8)=-15\\&f(13)=-14\\&\text{Although the total interval has a length of }20\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-7}^{13}f(x)\approx5[(-17)+(-16)+(-15)+(-14)]\\&\int_{-7}^{13}f(x)\approx-310\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(4)=9\\&f(6)=10\\&f(8)=-18\\&f(10)=15\\&f(12)=18\\&\text{Although the total interval has a length of }8\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{4}^{12}f(x)\approx2[(10)+(-18)+(15)+(18)]\\&\int_{4}^{12}f(x)\approx50\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=-13\\&f(6)=-4\\&f(14)=-15\\&f(22)=-19\\&f(30)=18\\&f(38)=-8\\&\text{Although the total interval has a length of }40\\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }8\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{38}f(x)\approx8[(-4)+(-15)+(-19)+(18)+(-8)]\\&\int_{-2}^{38}f(x)\approx-224\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(0)=-19\\&f(5)=14\\&f(10)=2\\&f(15)=15\\&\text{Although the total interval has a length of }15\\&\text{, notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{0}^{15}f(x)\approx5[(14)+(2)+(15)]\\&\int_{0}^{15}f(x)\approx155\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=7\\&f(-6)=-7\\&f(-4)=16\\&f(-2)=-16\\&f(0)=20\\&\text{Although the total interval has a length of }8\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-8}^{0}f(x)\approx2[(-7)+(16)+(-16)+(20)]\\&\int_{-8}^{0}f(x)\approx26\end{align*}\)