The distance covered by the object can be found by integrating the velocity equation from time $\displaystyle t= 2$ to time $\displaystyle t = 3$. To integrate the velocity equation we can use the power rule where if 

$\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Using this rule, the distance is calculated as

$\displaystyle x = \int_2^3 (3t^2 - 5)dt = \frac{3t^{(2+1)}}{2+1} - \frac{5t^{(0+1)}}{0+1}]_2^3 = t^3 - 5t]_2^3$

$\displaystyle x = (3)^3 - 5(3) - (2)^3 + 5(2) = 27 - 15 - 8 +10 = 14$

The distance travelled by the object can be found by integrating the velocity equation.

Because the derivative of $\displaystyle \sin(t)$ is $\displaystyle \cos(t)$, the integral of $\displaystyle \cos(t)$ is $\displaystyle \sin(t)$.

Therefore

$\displaystyle x = \int_0^1 (\frac{1}{2}\cos(t))dt = \frac{1}{2}\sin(t)]_0^1 = \frac{1}{2}\sin(1) - \frac{1}{2}\sin(0) = \frac{\sin(1)}{2} - 0$

$\displaystyle x = \frac{\sin(1)}{2}$

The distance travelled by the object can be found by integrating the position equation for the object from $\displaystyle t =1$ to $\displaystyle t =2$. This can be done using the power rule where if

$\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$.

Using the power rule the distance travelled by the object is

$\displaystyle s = \int_1^2 (2t - 6t^2)dt = \frac{2t^{(1+1)}}{1+1} - \frac{6t^{(2+1)}}{2+1}]_1^2 = t^2 - 2t^3]_0^1$.

$\displaystyle s = (2)^2 - 2(2)^3 - (1)^2 + 2(1)^3= 4 - 16 - 1 + 2 = -11$

The distance travelled can be found by integrating the velocity from $\displaystyle t= 0$ to $\displaystyle t =3$.

The velocity can be integrated using the power rule where

$\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$.

Applying this to the velocity equation gives

$\displaystyle \\s = \int_0^3 (4t -7t^0)dt = \frac{4t^{(1+1)}}{1+1} -\frac{7t^{(0+1)}}{0+1}]_0^3 = 2t^2 - 7t]_0^3 \Rightarrow \\ \\ \Rightarrow 2(3)^2 - 7(3) - 2(0)^2 + 7(0) = 18 - 21 - 0 + 0 = -3$

To find the distance between the two locations we can subtract the position at $\displaystyle t = 1$ from the position at $\displaystyle t = 3$.

$\displaystyle \\x(3) - x(1) = 2(3)^3 - 3(3) +4 - 2(1)^3 + 3(1) - 4 \\= 54 - 9 +4 -2 +3 -4 \\= 46$

Therefore the distance between the locations is $\displaystyle 46$.

To find the distance travelled we can integrate the velocity equation of the object.

This can be done using the power rule where if

$\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$.

Using this equation we find that, 

$\displaystyle s = \int_0^{10} (300t-20t^3+500t^0)dt = \frac{300t^{1+1}}{1+1} - \frac{20t^{3+1}}{3+1} +\frac{500t^{0+1}}{0+1}]_0^{10}$

$\displaystyle s = 150(10)^2 - 5(10)^4 + 500(10) -150(0)^2 - 5(0)^4 +500(0)$

$\displaystyle s = 15,000 - 50,000 +5,000 = -30,000$.

To find the distance covered by the object we can integrate the velocity equation. This can be done using the power rule where if

 $\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$.

Therefore the distance covered by the object is

$\displaystyle s = \int_0^2 (8t^3)dt = \frac{8t^{(3+1)}}{3+1}]_0^2 = 2t^4]_0^2 = 2(2)^4 - 2(0)^4 = 32$.

The distance covered by the object can be found by integrating the acceleration twice. This can be done using the power rule where if

$\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$.

Using this rule gives 

$\displaystyle v(t) = \int (6t)dt = \frac{6t^{(1+1)}}{1+1} + C = 3t^2 +C$.

The value of $\displaystyle C$ can be found using the initial velocity of the object.

$\displaystyle v(0) = 3(0)^2 + C = 0 +C = 3$

Therefore $\displaystyle C = 3$ and $\displaystyle v(t) = 3t^2 + 3$.

Integrating the velocity equation from $\displaystyle t= 0$ to $\displaystyle t= 1$ will give us the distance covered by the object.

$\displaystyle s= \int_0^1 (3t^2+3)dt = \frac{3t^{(2+1)}}{2+1} + \frac{3t^{(0+1)}}{0+1}]_0^1 = t^3 + 3t]_0^1$

$\displaystyle s = (1)^3 + 3(1) - (0)^3 - 3(0) = 1 + 3 -0 - 0= 4$

The distance covered by the object can be found by integrating the velocity from $\displaystyle t = 0$ to $\displaystyle t =1$, using the power rule where if

$\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$.

Applying this to the velocity equation gives

$\displaystyle s = \int_0^1 (30t^2-4)dt = \frac{30t^{2+1}}{2+1} - \frac{4t^{0+1}}{0+1}]_0^1 = 10t^3 - 4t]_0^1$

$\displaystyle s = 10(1)^3 - 4(1) - 10(0)^3 +4(0) = 10 -4 - 0 + 0 = 6$

The distance covered can be found by integrating the velocity from $\displaystyle t= 1$ to $\displaystyle t =2$ using the power rule, where if

$\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$.

Therefore the distance covered is

$\displaystyle s = \int_1^2 (20t - 3t^2)dt = \frac{20t^{(1+1)}}{1+1} - \frac{3t^{2+1}}{2+1}]_1^2 = 10t^2 - t^3]_1^2$

$\displaystyle s= 10(2)^2 - (2)^3 - 10(1)^2 + (1)^3 = 40 - 8 - 10 + 1 = 23$