Science:Math Exam Resources/Courses/MATH101/April 2018/Question 01 (i)

MATH101 April 2018

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Question 01 (i)
Evaluate ${\textstyle \displaystyle \int _{-2}^{2}\left(1-2\|x\|\right)\,dx}$ . A: ${\textstyle {2}}$ B: ${\textstyle {0}}$ C: ${\textstyle -4}$ D: ${\textstyle 4}$ E: ${\textstyle -2}$ F: ${\textstyle 1}$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Split the integral into ${\textstyle \int _{-2}^{2}=\int _{-2}^{0}+\int _{0}^{2}}$ to get rid of the absolute value ${\textstyle \|x\|}$ .

Hint 2
(Solution 2) Observe that the integrand is an even function.

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Solution 1
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. As suggested in the hint, we can write $\int _{-2}^{2}(1-2\|x\|)\;dx=\int _{-2}^{0}(1-2\|x\|)\;dx+\int _{0}^{2}(1-2\|x\|)\;dx.$ Then we have $\int _{-2}^{0}(1-2\|x\|)\;dx=\int _{-2}^{0}(1+2x)\;dx={\Big [}x+x^{2}{\Big ]}_{-2}^{0}=2-4=-2,$ and $\int _{0}^{2}(1-2\|x\|)\;dx=\int _{0}^{2}(1-2x)\;dx={\Big [}x-x^{2}{\Big ]}_{0}^{2}=2-4=-2.$ We conclude that the value of the original integral is ${\textstyle -4}$ . Answer: The correct answer is ${\textstyle {\color {blue}C}}$ .

Solution 2
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $f(x)=1-2\|x\|$ . Observe that it is an even function; $f(-x)=1-2\|-x\|=1-2\|x\|=f(x)$ . Therefore, we have $\int _{-2}^{2}1-2\|x\|dx=2\int _{0}^{2}1-2\|x\|dx.$ Evaluating the integral on the right hand side as in Solution 1, the given integral has the value $-4$ , so the answer is $C$ . Answer: $\color {blue}C$