Science:Math Exam Resources/Courses/MATH105/April 2017/Question 06 (a)

MATH105 April 2017

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Question 06 (a)
(a) Determine if there exist two continuous functions $f:[0,1]\to \mathbb {R}$ and $g:[0,1]\to \mathbb {R}$ such that $\int _{0}^{1}f(x)g(x)dx=0$, but $\int _{0}^{1}f(x)dx\neq 0$ and $\int _{0}^{1}g(x)dx\neq 0$. If your answer is YES, please give the two functions f and g explicitly by some formulas. If your answer is NO, explain why.

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 statement that $\int _{0}^{1}f(x)g(x)=0$ states that the product f(x) g(x) cancels out on average. Is it possible for this product to average out to zero without either of f or g averaging out to zero?

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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. One way to solve this problem is to set things up so that the function f(x) g(x) is equal to zero everywhere on the region of integration. For example, we could take $f(x)={\begin{cases}1-3x&{\text{if }}0\leq x\leq {\frac {1}{3}}\\0&{\text{if }}{\frac {1}{3}}<x\leq 1\end{cases}}$ and $g(x)={\begin{cases}0&{\text{if }}0\leq x\leq {\frac {2}{3}}\\3(x-2/3)&{\text{if }}{\frac {2}{3}}<x\leq 1\end{cases}}$ Then f(x)g(x) is equal to zero everywhere, but the region under each of f(x) and g(x) is a triangle of area ${\frac {1}{6}}$. Answer: $\color {blue}{\text{Yes, it is possible to find such functions.}}$