# How to solve this using definite integral?

I already have asked $2$ questions on similar topics. In that questions I have equations of parabola and line. And equations have $y$ and $x$ variable.

But in this question I have functions.

$$f(x) = |x| - 1 \quad\mbox{and}\quad g(x) = 1 - |x|.$$

a) Sketch their graphs.

b) Using integration find the area of the bounded region.

In other two questions I know how to find values of $x$ and $y$ using substitution. But in this question because of $f(x)$ and $g(x)$. I am clueless how to start.

My attempt is only.

$f(x) = x - 1, -x - 1$

$g(x) = 1 - x, 1 + x$.

• desmos.com/calculator/9sho5r7uue Dec 29, 2016 at 14:03
• But how to find value of y as we have no y variable in equation.
– user404716
Dec 29, 2016 at 14:05
• desmos.com/calculator/nlrqqou8bf Dec 29, 2016 at 14:06
• y=f (x) and y=g (x) Dec 29, 2016 at 14:14
• So you can write y=|x|-1 and y=1-|x| . Does that help ease the confusion or this is not what you meant about the confusion on f and g? Dec 29, 2016 at 14:16

I presumed you know how to sketch the region. Then $$A=2\int_{0}^1[(1-x)-(x-1)]dx=\int_0^1(4-4x)dx=2.$$

In set notation, the region $R$ is given by $$R=\{(x,y):-1\leq x\leq 1,f(x)\leq y\leq g(x)\}.$$ The line $x=0$ or known as the $y$-axis serves the line of symmetry.

Okay,this is the graph of the region $R$.

• @John Sr Feel free to ask--:D Dec 29, 2016 at 14:18
• Your answer is wrong. In book its 2. And I have main problem in drawing sketch. Because we have no y variable.
– user404716
Dec 29, 2016 at 14:18
• @John Sr Why is it wrong? Try to simplify my answer and the answer is 2. You want the solution? Dec 29, 2016 at 14:20
• @John Sr To find the points of intersection, we get $$1-|x|=|x|-1$$ and yields $|x|=1$ implying that $x=1$ or $x=-1$. Thus, their points of intersection are $(1,0)$ and $(-1,0)$. Dec 29, 2016 at 14:53
Hint for f (x) shift the graph of $|x|$ by 1 towards right. For other function use $g (x)=1+x$ when x is negative and $1-x$ for positive. Then find the intersection with $f (x)$and then the area
Normally when finding area we do $\int_{a}^{b} f(x)-g(x)\text{d}x$ where f is the "top" function along the interval. So looking at a sketch of the region g(x) is on top over the interval and the interval is from -1 to 1. So our integral is $\int_{-1}^{1}1-\left | x\right |-(\left | x \right |-1)\text{d}x=$ $\int_{-1}^{1} 2-2\left |x\right |\text{d}x$ intuitively we can see that $\int \left | x\right |\text{d}x=\frac{\left |x\right |x}{2}+C$ from the piecewise definition of $|x|$ so using this just makes the integral into a pretty standard definite integral. I personally prefer this way because you do not need to figure out where the sign changes and do multiple integrals. However this only works for absolute value functions in the form $\left |ax+b\right |$ with higher powers and other functions you do need multiple integrals. Hope this helps a bit