Science:Math Exam Resources/Courses/MATH105/April 2010/Question 01 (j)

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MATH105 April 2010

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Question 01 (j)
Use the trapezoidal rule with $n=3$ to approximate $\displaystyle \int _{-2}^{4}{\frac {dx}{1+\left({\frac {x}{2}}\right)^{4}}}$

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
What information do we need to do a trapezoidal rule approximation?

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

As the hint suggests, we should calculate all the information we need to begin. First, we find the width of the intervals: $\Delta x={\frac {b-a}{n}}={\frac {4-(-2)}{3}}=2$. We also need the endpoints of the intervals: $x_{k}=\displaystyle a+k\Delta x$, for k = 0 to k = 3. Thus the points are -2, 0, 2, and 4. Then the formula for the Trapezoidal rule is $T_{n}={\frac {\Delta x}{2}}(f(x_{0})+2f(x_{1})+2f(x_{2})+...+2f(x_{n-1})+f(x_{n}))$, so we can fill in our information: $T_{3}={\frac {2}{2}}\left({\frac {1}{1+\left({\frac {-2}{2}}\right)^{4}}}+2{\frac {1}{1+\left({\frac {0}{2}}\right)^{4}}}+2{\frac {1}{1+\left({\frac {2}{2}}\right)^{4}}}+{\frac {1}{1+\left({\frac {4}{2}}\right)^{4}}}\right)$, $T_{3}={\frac {1}{1+1}}+2{\frac {1}{1+0}}+2{\frac {1}{1+1}}+{\frac {1}{1+16}}$, $T_{3}={\frac {1}{2}}+2+1+{\frac {1}{17}}$, $T_{3}={\frac {121}{34}}\approx 3.5588$. Notice if we actually evaluate the integral, $\int _{-2}^{4}{\frac {1}{1+\left({\frac {x}{2}}\right)^{4}}}{\textrm {d}}x=3.8742$ so we see that the trapezoidal rule with only 4 points is already in fairly good agreement.

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