How to solve: $x^2 = \arctan(x)$ I'm asked to find the area bounded by the curves: $y = x^2$ and y$ = \arctan(x)$. To find the limits of integration, I set the two functions equal to each other and solve for $x$. Finding the left bound as 0 was trivial, but I'm unsure how to solve for the right bound.
$$x^2 = \arctan(x)$$

 A: An antiderivative for $\arctan x$ is
$$
x\arctan x-\frac{1}{2}\log(1+x^2)
$$
so the area you're asked to compute is
$$
\int_0^c (\arctan x-x^2)\,dx=
\Bigl[x\arctan x-\log(1+x^2)-\frac{x^3}{3}\Bigr]_0^c=
c^3-\frac{1}{2}\log(1+c^2)-\frac{c^3}{3}
$$
where $c^2=\arctan c$. Now compute $c$ with an approximation method, if you want an approximation of the result.
With $x_0=1$ and $x_{n+1}=\sqrt{\arctan x}$, I get $x_{10}=0.8336$, which should be exact to four decimals and the integral evaluates to $\approx0.122$.
A: We can't obtain a closed form for the not trivial solution and we need to proceed by numerical methods (Newton's, bisection, etc.).
Here is the numerical solution by WolframAlpha that is $x \approx 0.833606194406676$.
If you want to obtain the result by yourself let consider $f(x)=x^2-\arctan x$ and use bisection method by a calculator starting for example from


*

*$x_a=1 \implies f(1)>0$

*$x_b=0.8 \implies f(0.8)<0$
and then we can iteretively get closer and closer to the solution by


*

*$x_i=\frac{x_a+x_b}2$

*if $f(x_i)>0 \implies x_a=f(x_i)$ 

*if $f(x_i)<0 \implies x_b=f(x_i)$
A: You want to find the point of intersection of $ y=x^2$ and $y= \tan ^{-1} x $
With Newton's method and $$f(x) = x^2 - \tan^ {-1} x$$ you get a numerical scheme, $$ x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)} $$ which converges to the desired solution.
The starting point should be reasonable, such as $x=1$ 
A: Sometimes getting you in the ballpark can give you clues as to how to proceed. 
By Taylor Series:
$\frac{1}{1+x^2}=1-x^2+x^4-x^6+...$
It's integral is : $Tan^{-1}(x)= x-x^3/3+...$
So, solve $x^2=x-x^3/3$. Dividing by $x$ and rearranging:
$x^2/3+x-1=0.$
So by the quadratic formula, Let $x_0=\frac{-1\frac{+}{-}\sqrt{1+4/3}}{2/3}$
Using integration by parts:
$$\int_0^{x_0} Tan^{-1}x-x^2 dx=xTan^{-1}(x)|_0^{x_0} - \int_0^{x_0} \frac{x}{1+x^2}+x^2 dx$$
$$=\frac{2}{3}x_0^3-\frac{1}{2}ln(1+x_0^2)$$
You can tweak $x_0$ some more via Newton's Method or other approximation method. 
A: In addition to Taylor Series, you can use Banach's Fixed Point Theorem. 
It can be shown that if we have $|x-c|<\epsilon, |f'(x)|<1, $and$ f(x)=x $ then $|f(x)-f(c)|<\epsilon.$ 
So:
$$x^2=Tan^{-1}(x)$$
$$x=\sqrt{Tan^{-1}(x)}$$
Let $f(x)=\sqrt{Tan^{-1}(x)}$
Then $f'(x)=\frac{1}{2\sqrt{Tan^{-1}(x)}}\frac{1}{1+x^2}$
So $f'(x)=\frac{1}{2x+2x^3}$
We know that $x$ is between $.7$ and $1.0$. So regardless, the derivative is in the correct range. 
$$f(1)=\sqrt{\pi}/2=0.88623$$
$$f(\sqrt{\pi}/2)=0.85156$$
It's similar to Newton's method which would have you iterate:
$$x-\frac{x^2-Tan^{-1}(x)}{2x-\frac{1}{1+x^2}}$$
So less computational work but somewhat slower speed of convergence.
