Science:Math Exam Resources/Courses/MATH101/April 2005/Question 07 (c)

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MATH101 April 2005

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8 (a)  •  Q8 (b)  •  

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Question 07 (c)
The length of time in minutes it takes students to solve a certain mathematics problem (on probability) is a continuous random variable whose probability density function is $f(x)={\begin{cases}k\cos(x/10)\sin ^{2}(x/10),&{\text{if }}0\leq x\leq 5\pi ,\\0,&{\text{otherwise.}}\end{cases}}$ (c) Compute the mean length of time required to solve the problem.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The mean of a random variable X with probability density function f(x) is given by $\int _{-\infty }^{\infty }xf(x)\,dx.$

Hint 2
You will need integration by parts to calculate this integral.

Hint 3
Choose u = x, dv = 3/10 cos(x/10) sin2(x/10). From your calculations in part (a) and (b), how can you quickly find v?

Hint 4
Once your integration by parts is all done you should find the integral $\int _{0}^{5\pi }\sin ^{3}\left({\frac {x}{10}}\right)\,dx$ Use the trigonometric Pythagoras to solve this integral and finish the problem.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

From the definition of the mean we know that we have to calculate the mean μ by $\mu =\int _{0}^{5\pi }x{\frac {3}{10}}\cos \left({\frac {x}{10}}\right)\sin ^{2}\left({\frac {x}{10}}\right)\,dx.$ As suggested by the hint we choose u = x, du = dx, dv = 3/10 cos(x/10)sin2(x/10), v = ?? To find v here we use the same procedure that we already used in part (a) and part (b), i.e. the substitution y = sin(x/10) (let's use y instead of u, since u is already in use at the integration by parts) to find ${\begin{aligned}\int {\frac {3}{10}}\cos \left({\frac {x}{10}}\right)\sin ^{2}\left({\frac {x}{10}}\right)\,dx&=\int 3y^{2}\,dy\\&=y^{3}+C\\&=\sin ^{3}\left({\frac {x}{10}}\right)+C.\end{aligned}}$ With this intermediate result we can now go to work at calculating the mean μ: ${\begin{aligned}\mu &=\int _{0}^{5\pi }\underbrace {x} _{=u}\underbrace {{\frac {3}{10}}\cos \left({\frac {x}{10}}\right)\sin ^{2}\left({\frac {x}{10}}\right)\,dx} _{=dv}\\&=\left.x\sin ^{3}\left({\frac {x}{10}}\right)\right|_{0}^{5\pi }-\int _{0}^{5\pi }\sin ^{3}\left({\frac {x}{10}}\right)\,dx\\&=5\pi -\int _{0}^{5\pi }\sin ^{3}\left({\frac {x}{10}}\right)\,dx\end{aligned}}$ Let's compute the last integral separately, and only then come back to this equation. To do so, we need the trigonometric identity $\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1.$ With this we obtain ${\begin{aligned}\int _{0}^{5\pi }\sin ^{3}\left({\frac {x}{10}}\right)\,dx&=\int _{0}^{5\pi }\sin \left({\frac {x}{10}}\right)\left(1-\cos ^{2}\left({\frac {x}{10}}\right)\right)\,dx\\&=\int _{0}^{5\pi }\sin \left({\frac {x}{10}}\right)\,dx+\int _{0}^{5\pi }-\sin \left({\frac {x}{10}}\right)\cos ^{2}\left({\frac {x}{10}}\right)\,dx\end{aligned}}$ The former integral is standard, for the latter we substitute z = cos(x/10), dz = -1/10 sin(x/10)dx, z(0) = cos(0) = 1, z(5π) = cos(π/2) = 1. Hence ${\begin{aligned}\int _{0}^{5\pi }\sin ^{3}\left({\frac {x}{10}}\right)\,dx&=\int _{0}^{5\pi }\sin \left({\frac {x}{10}}\right)\,dx+\int _{0}^{5\pi }-\sin \left({\frac {x}{10}}\right)\cos ^{2}\left({\frac {x}{10}}\right)\,dx\\&=\left.-10\cos \left({\frac {x}{10}}\right)\right|_{0}^{5\pi }+\int _{1}^{0}10z^{2}\,dz\\&=0+10+\left.{\frac {10}{3}}z^{3}\right|_{1}^{0}\\&=10+0-{\frac {10}{3}}\\&={\frac {20}{3}}.\end{aligned}}$ Alright. What did we want to do with the 20/3? Right, we needed to calculate this value to find μ. Going back to the equation with μ we can now completely the question by plugging in the result: ${\begin{aligned}\mu &=\int _{0}^{5\pi }x{\frac {3}{10}}\cos \left({\frac {x}{10}}\right)\sin ^{2}\left({\frac {x}{10}}\right)\,dx\\&=5\pi -\int _{0}^{5\pi }\sin ^{3}\left({\frac {x}{10}}\right)\,dx\\&=5\pi -{\frac {20}{3}}.\end{aligned}}$

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