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

MATH101 April 2010

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Question 05 (a)
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: $\ f(0)=1,$ $\ f(1)=2,$ $\ f(2)=0,$ $\ f(3)=-2,$ $\ f(4)=-4.$ Write down the trapezoidal approximation $T_{4}$ for $\int _{0}^{4}f(x)dx.$

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 formula for the trapezoid rule is given by $T_{n}:={\frac {\Delta x}{2}}\left(f(x_{0})+2f(x_{1})+2f(x_{2})+...+2f(x_{n-1})+f(x_{n})\right)$

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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. Since the formula for the trapezoid rule can be given by $T_{n}={\frac {\Delta x}{2}}\left(f(x_{0})+2f(x_{1})+2f(x_{2})+...+2f(x_{n-1})+f(x_{n})\right)$ and we know that we want to evaluate this rule using $n=4$, we have $x_{i}=a+i\Delta x=0+i{\frac {4-0}{4}}$ and thus, the trapezoidal approximation that we need is ${\begin{aligned}T_{4}&={\frac {1}{2}}\left(f(0)+2f(1)+2f(2)+2f(3)+f(4)\right)\\&={\frac {1}{2}}(1+2(2)+2(0)+2(-2)+(-4))={\frac {-3}{2}}\end{aligned}}$

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