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

MATH101 April 2010

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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.

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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.

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