Respuesta :
Answer:
(A). 3.93
Step-by-step explanation:
You need to find the following integral
[tex]{\displaystyle \int\limits_{2}^{6} 4(t-6)\cos(2t+5) \, dt } = 3.93[/tex]
Therefore the answer is (A). 3.93
The total distance traveled by the referee from [tex]t = 2\,min[/tex] and [tex]t = 6\,min[/tex] is approximately 9.814 meters.
How to determine total distance of the referee by definite integrals
Physically speaking, the total distance ([tex]s[/tex]), in meters, is the integral of velocity ([tex]v(t)[/tex]), in meters per minute, in a given interval of time ([tex]t[/tex]), in minutes. Hence, we must solve the following integral to calculate the total distance traveled by the referee:
[tex]s = \int\limits^{6}_{2} {4\cdot (t-6)\cdot \cos (2\cdot t + 5)} \, dt[/tex] (1)
Now we proceed to solve the definite integral:
[tex]s = 4\int\limits^{6}_{2} {t\cdot \cos (2\cdot t+5)} \, dt -24\int\limits^{6}_{2} {\cos (2\cdot t +5)} \, dt [/tex] (1b)
By the algebraic substitution [tex]u = 2\cdot t + 5[/tex] ([tex]du = 2\cdot dt[/tex]) we have the expression:
[tex]s = 2\int\limits^{6}_{2} {\left(\frac{u-2}{5} \right)\cdot \cos u} \, du - 12 \int\limits^{6}_{2} {\cos u} \, du [/tex]
[tex]s = \frac{2}{5}\int\limits^{6}_{2} {u\cdot \cos u} \, du - \frac{4}{5} \int\limits^{6}_{2} {\cos u} \, du - 12\int\limits^{6}_{2} {\cos u} \, du [/tex]
[tex]s = \frac{2}{5}\int\limits^{6}_{2} {u\cdot \cos u} \, du -\frac{64}{5}\int\limits^{6}_{2} {\cos u} \, du [/tex]
The solution of this integral is:
[tex]s = \frac{2}{5}\cdot (u\cdot \sin u + \cos u)|_{2}^{6}-\frac{64\cdot \sin u}{5}|_{2}^{6} [/tex]
[tex]s = \frac{2}{5}\cdot [(2\cdot t + 5)\cdot \sin (2\cdot t + 5) + \cos (2\cdot t + 5)]|_{2}^{6} - \frac{64}{5}\cdot \sin (2\cdot t + 5)|_{2}^{6} [/tex]
[tex]s \approx 9.814\,m[/tex]
The total distance traveled by the referee from [tex]t = 2\,min[/tex] and [tex]t = 6\,min[/tex] is approximately 9.814 meters. [tex]\blacksquare[/tex]
Remark
The statement is poorly formatted and presents mistakes, correct form is shown below:
A referee moves along a straight path on the side of an athletic field. The velocity of the referee is given by [tex]v(t) = 4\cdot (t-6)\cdot \cos (2t+5)[/tex], where [tex]t[/tex] is measured in minutes and [tex]v(t) [/tex] is measured in meters per minute.
What is the total distance traveled by the referee, in meters, from [tex]t= 2\,min[/tex] to [tex]t = 6\,min[/tex]?
To learn more on definite integrals, we kindly invite to check this verified question: https://brainly.com/question/22655212
