# $\int _ {- 1} ^ {3} [| x ^ 2-6x | + \text{sgn} (x-2)]\, \text {d} x$

Calculate the integral $$\int _ {- 1} ^ {3} \bigl[| x ^ 2-6x | + \text{sgn} (x-2)\bigr]\, \text {d} x .$$

We know that the two functions are defined as follows, \begin{align*} f (x) &= | x ^ 2-6x |\\[5pt] &=\begin{cases} x^2-6x, & \text{ if } x^2-6x>0; \\ 6x-x^2, & \text{ if } x^2-6x<0. \end{cases}\\[5pt] &=\begin{cases} x^2-6x, & \text{ if } x\in(-\infty, 0)\cup (6,+\infty); \\ 6x-x^2, & \text{ if } x\in(0,6). \end{cases}. \end{align*} And, $$\text{sgn}(x-2)=\begin{cases} 1, & \text{ if } x>2; \\ -1, & \text{ if } x<2. \end{cases}$$

I know that the integral gives $$58 /$$ 3. But someone can explain to me the step by step, how to find that result.

I considered doing it with the graph of the function which is the following, Calculating, the area of ​​the triangle on the left from $$-1$$ to $$0$$, and then calculating the two on the right and adding it, but I don't know how to calculate the area on the right.

• Illuminating would be to graph and calculate $\int |x^2-6x|\,dx$ and $\int \text{sgn}(x-2)\,dx$ separately. Mar 16, 2021 at 17:49
• Yeah, you right. Typographical error. Thanks. Mar 16, 2021 at 17:57

You have all the necessary details in your explanation. It's just a matter of cleverly splitting the integral. $$\int_{-1}^3 [|x^2-6x| + \operatorname{sgn}(x-2)]dx = \int_{-1}^0 [x^2 - 6x -1] dx + \int_0^2 [6x - x^2 -1]dx + \int_2^3[6x - x^2 +1] dx$$ Now you can integrate the three integrals separately

General strategy Say you are trying to integrate a piecewise function on $$[a,b]$$ $$f = \begin{cases} f_1 & x \in (x_1, x_2) \\ f_2 & x \in (x_2, x_3) \\ \vdots \\ f_n & x \in (x_n, x_{n+1})\end{cases}$$ It helps to split the integral in the following fashion $$\int_a^b f = \int_{x_1}^{x_2} f_1 + \dots + \int_{x_n}^{x_{n+1}}f_n$$
In your case, the function you are trying to integrate was over $$[-1,3]$$ and the pieces, like you figured out were $$f = |x^2 - 6x|+\operatorname{sgn}(x-2) = \begin{cases} x^2-6-1 & x \in (-1,0) \\ 6x-x^2-1 & x\in (0,2) \\ 6x-x^2 +1 & x\in (2,3) \end{cases}$$

• Thanks, for help me. :) Mar 16, 2021 at 17:58

Let's first work on getting a piecewise equation for the whole thing. There are three points where the pieces can change, namely $$x = 0, 2, 6$$. So, we will consider each interval separately when writing down the piecewise equation. For example, for $$x<0$$, we need to consider the following: $$\begin{cases} \color{red}{x^2-6x}, & \color{red}{x\in(-\infty, 0)\cup (6,+\infty)} \\ 6x-x^2, & x\in(0,6) \end{cases} + \begin{cases} 1, & x>2; \\ \color{red}{-1}, & \color{red}{x<2}. \end{cases} = \boxed{x^2 - 6x - 1,\quad x<0 \ }$$

In the same way we can obtain the rest of the cases: $$\boxed{| x ^ 2-6x | + \operatorname{sgn} (x-2) = \begin{cases} x^2 - 6x -1, & x<0\\ 6x - x^2 -1, &02 \end{cases}}$$ So, this tells us that the integral can be split up as $$\boxed{\int_{-1}^{0}\bigr[ x^2 - 6x - 1\bigl]\, \text {d}x + \int_{0}^{2}\bigl[6x - x^2 -1\bigr]\, \text {d}x + \int_{2}^{3}\bigl[6x - x^2 + 1\bigr] \, \text {d}x \ }$$

Basic strategy in these type of problems involving Piece-wise functions is that Firstly you have to look at Limits of your integral and then check if Point of discontinuity(Here $$0$$ ,$$2$$ and $$6$$) lies in limit of your integral. If They lies You have to use Linearity of integral.

$$I=\int _ {- 1} ^ {3} \bigl[| x ^ 2-6x | + \text{sgn} (x-2)\bigr]\, \text {d} x$$

You have already determine Points where we have to use linearity of integral.

$$I=\int _ {- 1} ^ {0} (x ^ 2-6x -1 )dx + \int _ {0} ^ {2} (-x ^ 2+6x -1) dx +\int _ {2} ^ {3} (-x ^ 2+6x +1) dx$$

Now, You can evaluate integrals easily.