Science:Math Exam Resources/Courses/MATH101/April 2014/Question 02/Solution 1

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 ${\frac {8}{\pi }}-1$.