**Answer:**

a) 103.176 m / min

b) 1751.28 meters

**Explanation:**

**Given:-**

- Emma's and Lily's velocities ( E(t) and L(t) ) are given as functions respectively:

[tex]E(t) = \frac{7510}{t^2-7t + 80.22} \\\\L ( t ) = 12t^3*e^-^0^.^5^t[/tex]

- Where, E ( t ) and L ( t ) are given in m / min

- Both run for a total time of 15 minutes in the same direction along the straight track defined by the absolute interval:

( 0 ≤ t ≤ 15 ) mins

- It is known that Emma is 10 meters ahead of Lily at time t = 0.

**Find:-**

a) Find the value of [tex]\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt[/tex] using correct units, interpret the meaning of

b) What is the maximum distance between Emma and Lily over the time interval 0 ≤ t ≤ 15?

**Solution:-**

- The average value of a function f ( x ) over an interval [ a , b ] is determined using calculus via the following relation:

[tex]f_a_v_g = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx[/tex]

- The first part of the question is asking us to determine the average velocity of Emma over the time interval of ( 2 , 8 ). Therefore, ( E_avg ) can be determined using the above relation:

[tex]E_a_v_g = \frac{1}{8 - 2}*\int\limits^8_2 {E(t)} \, dt\\\\E_a_v_g = \frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\\[/tex]

- We will evaluate the integral formulation above to determine Emma's average velocity over the 2 ≤ t ≤ 8 minute range:

[tex]E_a_v_g = \frac{1}{6}*\int\limits^8_2 {\frac{7510}{t^2 - 7t + 80.22} } \, dt\\\\E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{50t^2 - 350t + 4011} } \, dt\\\\[/tex]

- Complete the square in the denominator:

[tex]E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{(5\sqrt{2}*t - \frac{35}{\sqrt{2} })^2 + \frac{6797}{2} } } \, dt\\\\[/tex]

- Use the following substitution:

[tex]u = \frac{5*(2t - 7 )}{\sqrt{6797} } \\\\\frac{du}{dt} = \frac{10}{\sqrt{6797} } \\\\dt = \frac{\sqrt{6797}}{10}.du[/tex]

- Substitute the relations for (u) and (dt) in the above E_avg expression.

[tex]E_a_v_g = \frac{1}{6}*37550\int {\frac{\sqrt{6797} }{5*(6797u^2 + 67997) } } \, du\\\\E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}} \int {\frac{1 }{(u^2 + 1) } } \, du[/tex]

- Use the following standard integral:

[tex]arctan(u) = \int {\frac{1}{u^2 + 1} } \, du[/tex]

- Evaluate:

[tex]E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}}* arctan ( u ) |[/tex]

- Apply back substitution for ( u ):

[tex]E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\[/tex]

- Plug in the limits and find Emma's average velocity:

[tex]E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}[/tex]

**Answer: Emma's average speed over the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.**

- The displacement S ( E ) of Emma from time t = 0 till time ( t ) over the absolute interval of 0≤t≤15 is given by the relation:

[tex]S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\[/tex]

- The displacement S ( L ) of Lily from time t = 0 till time ( t ) over the absolute interval of 0 ≤ t ≤ 15 is given by the relation:

[tex]S (L) = \int\limits^t_0 {L(t)} \, dt\\\\S (L) = \int\limits^t_0 ({12t^3 *e^-^0^.^5^t } )\, .dt\\[/tex]

Apply integration by parts:

[tex]S ( L ) = 24t^3*e^-^0^.^5^t - 64*\int\limits^t_0 ({e^-^0^.^5^t*t^2} \,) dt\\[/tex]

Re-apply integration by parts 2 more times:

[tex]S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2\int\limits^t_0 ({e^-^0^.^5^t*t} \,) dt ]\\[/tex] [tex]S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2*( -2t*e^-^0^.^5^t - (4e^-^0^.^5^t - 4 ) ]\\\\[/tex]

[tex]S ( L ) = e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) - 512 \\[/tex]

- The distance between Emma and Lily over the time interval 0 < t < 15 mins can be determined by subtracting S ( L ) from S ( E ):

[tex]S = S ( E ) - S ( L )\\\\S = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) + 887.71098\\[/tex]

- The maximum distance ( S ) between Emma and Lily is governed by the critical value of S ( t ) for which function takes either a minima or maxima.

- To determine the critical values of the function S ( t ) we will take the first derivative of the function S with respect to t and set it to zero:

[tex]\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671[/tex]

- We will plug in each value of t and evaluate the displacement function S(t) for each critical value: