# Solving $\int^b_0x=\int^b_0\cos x$

$$R$$ is the region below the curve $$y=x$$ and above the $$x$$-axis from $$x=0$$ to $$x=b$$, where $$b$$ is a positive constant. $$S$$ is the region below the curve $$y=\cos x$$ and above the $$x-axis$$ from $$x=0$$ to $$x=b$$. For what value of $$b$$ is the area of $$R$$ equal to the area of $$S$$?

This is the question I was given. I took this and set up an equation: $$\int^b_0x=\int^b_0\cos x$$I then took the integrals getting: $$\frac{b^2}{2}=\sin b$$ I solved for $$b$$ and got $$b=\sqrt{2\sin b}$$ The question given to me had multiple choice answers, so I just plugged them all in until I got the right answer - $$1.404$$. The other multiple choice options were: $$.739$$$$.877$$$$.986$$$$4.712$$
However, I am left with a question:Is there any way to solve this without being given multiple choice options?

• Only with numerical methods - bisection is my favorite, as it's more stable than many other methods, as well as guaranteed to converge if you can bracket a solution with a continuous function. – Adrian Keister Aug 2 '19 at 15:44
• Looks like you need numerical methods for this. However, in many multiple choice questions, you can make an educated guess about the correct answer, so you don't need to plug-in all the values. Can you mention the answer choices as well? – jgsmath Aug 2 '19 at 16:17
• just did it. see edit – Burt Aug 2 '19 at 16:21
• You might realize that $(\pi/2)^2/2\approx1.2$ and $\sin(\pi/2)=1$, so a solution is likely near $\pi/2\approx1.57$. To get a better estimate, you could then Taylor expand $\sin(b)$ at $b=\pi/2$ to a low order and solve the resulting polynomial equation. – MathIsFun7225 Aug 2 '19 at 16:36

If you had a calculator to check you answer, you had one to solve the problem with some numerical method. The most basic and simple one would be Fixed Point Iteration. You'd get a good enough answer in 4 steps. Particularly, this is easy if your calculator allows you to write sqrt(2*sin(ANS)) and press enter repeatedly to keep iterating.
$$\begin{matrix} b & \sqrt{2\sin(b)}\\ 1 &1,297282533\\ 1,297282533 &1,387679869\\ 1,387679869 &1,402341597\\ 1,402341597 &1,404168809\\ 1,404168809 &1,404385791\\ 1,404385791 &1,404411400\\ 1,404411400 &1,404414420\\ 1,404414420 &1,404414776\\ 1,404414776 &1,404414818\\ 1,404414818 &1,404414823 \end{matrix}$$
As MathIsFun commented, graphing or by inspection you notice that the solution is close to $$\frac \pi 2$$. Building the simplest Taylor series, you would get $$\frac{b^2}{2}-\sin b=\left(\frac{\pi ^2}{8}-1\right)+\frac{1}{2} \pi \left(b-\frac{\pi }{2}\right)+O\left(\left(b-\frac{\pi }{2}\right)^2\right)$$ giving as a very first approximation $$b_{(1)}=\frac{8+\pi ^2}{4 \pi }\approx 1.42202$$ Using one more term for the series expansion $$\frac{b^2}{2}-\sin b=\left(\frac{\pi ^2}{8}-1\right)+\frac{1}{2} \pi \left(b-\frac{\pi }{2}\right)+\left(b-\frac{\pi }{2}\right)^2+O\left(\left(b-\frac{\pi }{2}\right)^4\right)$$ giving as a second approximation $$b_{(2)}=\frac{1}{4} \left(\pi +\sqrt{16-\pi ^2}\right)\approx 1.40439$$ For sure, you could use Newton method which, starting with $$b_0=\frac \pi 2$$ would generate the following iterates $$\left( \begin{array}{cc} n & b_n \\ 0 & 1.570796327 \\ 1 & 1.422017936 \\ 2 & 1.404656638 \\ 3 & 1.404414871 \\ 4 & 1.404414824 \end{array} \right)$$
To get "nice looking" approximations, instead of Taylor series, you could use $$[1,n]$$ Padé approximants which would write $$\frac{b^2}{2}-\sin b\sim\frac {\left(\frac {\pi^2}8-1\right)+a_1^{(n)}\left(b-\frac{\pi }{2}\right)} {1+\sum_{p=1}^n c_p^{(n)}\left(b-\frac{\pi }{2}\right)^p}$$ giving as estimates $$b_{(n)}=\frac \pi 2- \frac{\frac {\pi^2}8-1 }{a_1^{(n)}}$$ For example, using $$n=2$$ would give $$b_{(2)}=\frac{64+32 \pi ^2-\pi ^4}{64 \pi }\approx 1.40444$$