Science:Math Exam Resources/Courses/MATH101/April 2010/Question 05 (c)

MATH101 April 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6  •  Q7  •  Q8  •  Q9  •  

   Other MATH101 Exams

  •  April 2011  •  April 2010  •  April 2009  •  April 2008  •  April 2007  •  April 2005  •  April 2006  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2018  •  April 2017  •  

Question 05 (c)

Full-Solution Problem. Justify your answer and show all your work. Simplification of the answer is not required. An unknown function $f(x)$ has the following values: $f(0)=1$, $f(1)=2$, $f(2)=0$, $f(3)=-2$, $f(4)=-4.$ It is known that the fourth derivative $f^{(4)}(x)$ lies between $-3$ and $2$ on the interval $[0,4]$. What is the largest possible value that $\int _{0}^{4}f(x)dx$ could have? You may use the fact that if $\left|f^{(4)}(x)\right|\leq K$ on the interval $[a,b]$, then the error in using $S_{n}$ to approximate $\int _{a}^{b}f(x)dx$ has absolute value less than or equal to $K(b-a)^{5}/180n^{4}$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The previous part of this question gives us the value of the Simpson's rule approximation. Find the worst value the error can be and then sum this worst value with the approximation from the previous part.

Checking a solution serves two purposes: helping you if, after having used the hint, 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Using the hint, we plug in $n=4$ into the error bound and notice that the error lies between ${\frac {(-3)(4-0)^{5}}{180(4)^{4}}}<{\frac {K(b-a)^{5}}{180n^{4}}}<{\frac {(2)(4-0)^{5}}{180(4)^{4}}}$ Evaluating these bounds, we get that ${\frac {-12}{180}}<{\frac {K(b-a)^{5}}{180n^{4}}}<{\frac {8}{180}}$ and one last simplification yields ${\frac {-1}{15}}<{\frac {K(b-a)^{5}}{180n^{4}}}<{\frac {2}{45}}$. Now, let's use the the approximation and notice that by definition, we have that $\left|\int _{0}^{4}f(x)\,dx-S_{4}\right|<{\frac {K(b-a)^{5}}{180n^{4}}}$ Notice that in magnitude, the value ${\frac {1}{15}}>{\frac {2}{45}}$ and so we have that the largest the integral can be is the $S_{4}$ approximation added onto this largest error bound. This gives $\int _{0}^{4}f(x)\,dx<S_{4}+{\frac {1}{15}}=-1+{\frac {1}{15}}={\frac {-14}{15}}$ completing the question.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Simpson's rule, Pages using DynamicPageList parser function, Pages using DynamicPageList parser tag