Science:Math Exam Resources/Courses/MATH101/April 2014/Question 02

MATH101 April 2014

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Question 02
Long Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Find the area of the finite region bounded between the two curves $\displaystyle y={\sqrt {2}}\cos(\pi x/4)$ and $\displaystyle y=|x|$. It will be useful to sketch the region first.

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
Where do the functions intersect? How does that help define the regions?

Hint 2
How does the area for positive $x$ values compare to the area for negative $x$ values.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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 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. The given region is the following: According to the picture above, the region has symmetry with respect to the $y$-axis (i.e. it is the same on the left and right). Thus, it’s enough to compute the area when $x\geq 0$ or $x<0$ and then double it. We will solve for $x>0$. We have to find the boundary of the region which is where the curves intersect. Since we will find the area when $x\geq 0$, it’s enough to find the intersection point of two graphs on the first quadrant: $y=x$ and $y={\sqrt {2}}\cos({\frac {\pi x}{4}})$. ${\begin{aligned}x=&{\sqrt {2}}\cos \left({\frac {\pi x}{4}}\right)\\{\frac {x}{\sqrt {2}}}=&\cos \left({\frac {\pi x}{4}}\right)\end{aligned}}$ This may look difficult to solve by hand and in general it is. However, we have a few special values of cosine that we know and one of them is $\pi /4$ so it seems reasonable to check if $x=1$ works. Substituting $x=1$, ${\frac {1}{\sqrt {2}}}=\cos({\frac {\pi }{4}})$, so the equality holds and our guess works. Thus, the two graphs on the first quadrant intersects at $x=1$ and the region of interest is on $0\leq x\leq 1$. Compute the area. Denote the area of the region when $x\geq 0$ by $A_{+}$. Then since the trig function takes on larger values over the region, we have ${\begin{aligned}A_{+}=\int _{0}^{1}{\sqrt {2}}\cos({\frac {\pi x}{4}})-xdx=&\left[{\frac {4{\sqrt {2}}}{\pi }}\sin({\frac {\pi x}{4}})-{\frac {x^{2}}{2}}\right]_{x=0}^{x=1}\\=&{\frac {4}{\pi }}\cdot {\sqrt {2}}\cdot {\frac {1}{\sqrt {2}}}-{\frac {1}{2}}\\=&{\frac {4}{\pi }}-{\frac {1}{2}}.\end{aligned}}$ Since the area of the whole region in the problem is the double of $A_{+}$ (by the reasons discussed above), the answer is ${\frac {8}{\pi }}-1$.

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. This solution doesn’t use the symmetry of the region. Find the intersection points. Since we have ${\begin{aligned}|x|={\sqrt {2}}\cos({\frac {\pi x}{4}})\Leftrightarrow {\begin{cases}x={\sqrt {2}}\cos({\frac {\pi x}{4}})&x\geq 0\\-x={\sqrt {2}}\cos({\frac {\pi x}{4}})&x<0\end{cases}}\Leftrightarrow {\begin{cases}{\frac {x}{\sqrt {2}}}=\cos({\frac {\pi x}{4}})&x\geq 0\\-{\frac {x}{\sqrt {2}}}=\cos({\frac {\pi x}{4}})&x<0\end{cases}}\end{aligned}}$ by inspection, at $x=1$, ${\frac {1}{\sqrt {2}}}=\cos({\frac {\pi }{4}})$, so the first equality holds. Also, at $x=-1$, $-{\frac {-1}{\sqrt {2}}}={\frac {1}{\sqrt {2}}}=\cos({\frac {\pi }{4}})$, so the second equality holds. Thus, the two graph intersect at $x=\pm 1$. Compute the area of the region. ${\begin{aligned}{\text{Area}}&=\int _{-1}^{1}{\sqrt {2}}\cos({\frac {\pi x}{4}})-|x|dx\\&=\int _{-1}^{0}{\sqrt {2}}\cos({\frac {\pi x}{4}})-|x|dx+\int _{0}^{1}{\sqrt {2}}\cos({\frac {\pi x}{4}})-|x|dx\\&=\int _{-1}^{0}{\sqrt {2}}\cos({\frac {\pi x}{4}})+xdx+\int _{0}^{1}{\sqrt {2}}\cos({\frac {\pi x}{4}})-xdx\\&=\left[{\frac {4{\sqrt {2}}}{\pi }}\sin({\frac {\pi x}{4}})+{\frac {x^{2}}{2}}\right]_{x=-1}^{x=0}+\left[{\frac {4{\sqrt {2}}}{\pi }}\sin({\frac {\pi x}{4}})-{\frac {x^{2}}{2}}\right]_{x=0}^{x=1}\\&={\frac {4}{\pi }}\cdot {\sqrt {2}}\cdot {\frac {1}{\sqrt {2}}}-{\frac {1}{2}}+{\frac {4}{\pi }}\cdot {\sqrt {2}}\cdot {\frac {1}{\sqrt {2}}}-{\frac {1}{2}}\\&={\frac {8}{\pi }}-1.\end{aligned}}$

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