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

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

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