# Area of the region bounded by 2 curves

So I understand how to interpret a graph to determine the intersection of curves, however, this is the first time I've encountered an $y=|x|$.

Here is the problem, and my work. Can someone explain the absolute value concept in this regard?

Find the area of the region determined by the intersection of the curves. $y=\frac2{x^2+1}, y=|x|$.

$$A=\int_{-1}^1 (\frac2{x^2+1})dx$$

$$=2\int(\frac1{x^2+1})dx = 2 arctan x$$

$$\int x dx=\frac{x^2}2$$

$$A=\int_{-1}^1 (\frac2{x^2+1})dx=2 arctan x - \frac{x^2}2+C$$

$$\lim_{x \to -1+}{(2 arctan x - \frac{x^2}2)}=-2.071$$

$$\lim_{x \to 1-}{(2 arctan x - \frac{x^2}2)}=1.071$$

$$A\approx1.071-(-2.071)=3.142\approx\pi$$

Well, the problem is the same as it would be for some other function, except that $\lvert x\rvert=f(x)$ is a piecewise defined function, i.e. $$f(x)=\begin{cases} x&x\ge 0\\ -x&x<0. \end{cases}$$ Now, all we need to do is compute the integral $\int_{-1}^1\frac{2}{x^2+1}dx$ and subtract off the area lying above $f(x)=\lvert x\rvert$. There are a few ways to do this. The best way is to observe that $\lvert x\rvert$ is symmetric about the origin so that you can just take $$2\int_0^1 xdx=\int_{-1}^1\lvert x\rvert dx.$$ Another nice way to do this is with basic Euclidean geometry. We know that the area of the triangle bounded by $y=x, x=0,y=0$ has area $\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$. This corresponds to $\int_0^1 xdx$. Doubling this, we find that the area $\int_{-1}^1\lvert x\rvert dx=1.$ In any case, when all is said and done, we get that $$A=2\arctan(x)\bigg|_{-1}^1-1=2\arctan(1)-2\arctan(-1)-1=2\cdot \frac{\pi}{4}+2\frac{\pi}{4}-1=\pi-1.$$

i would write $$A=2\int_{0}^1\frac{2}{x^2+1}-xdx$$

Do you remember what the absolute value function really looks like? Well, mathematically it's defined like this:

$$|x|=\begin{cases} x & x ≥ 0\\ -x & x < 0 \end{cases}$$

When you have an absolute value function as part of your problem, you need to replace the original function with the equivalent functions on the respective intervals. On the interval $0≤x<\infty$, the function $|x|$ is positive which leaves you with the curve $f(x)=x$ on that interval. Therefore, $\frac2{x^2+1}$ is our upper curve and $x$ is our lower curve on the interval $0≤x<\infty$ (here's an entire lecture for you if you're not not sure how the idea behind finding the area between two curves works). And the point $(1,1)$ is where they cross on that interval. But also notice that the region formed by the curves $\frac2{x^2+1}$ and $x$ on the interval $0≤x<\infty$ is symmetric to the region on the left-hand side with respect to the y-axis which means that the two regions have the same area. So, all you have to do is to find the definite integral of $\frac2{x^2+1}-x$ from $0$ to $1$ (instead of from $-1$ to $1$) and double it:

\begin{align} A=2\int_{0}^{1} \left(\frac2{x^2+1}-x\right)dx &=4\int_{0}^{1}\frac{1}{x^2+1}dx-2\int_{0}^{1}xdx\\ &=4\cdot\arctan{x}\Big|_{0}^{1}-2\cdot\frac{x^2}{2}\Big|_{0}^{1}\\ &=4\cdot\left(\arctan{1}-\arctan{0}\right)-\left(1^2-0^2\right)\\ &=4\cdot\left(\frac{\pi}{4}-0\right)-1\\ &=\pi-1\\ \end{align}

• Thank you so much for the explanation. I see now looking again the symmetry that I missed before. Thanks again for the assist. – xerxes2985 Jan 12 '18 at 3:11