sketch the graph of the integrand function and use it to help evaluate the integral.
integration from(1 , -1) |x|-1
I think I can evaluate the integration
f(x) = 1/2 x^2 -x+c
but how sketch the graph
$\endgroup$ 03 Answers
$\begingroup$Another method would be to use the graph alone. Created using Desmos
From the graph, it is clear this is a triangle with base length $2$ and height $-1$. Since we know the integral is the area under the curve, and the area of a triangle is $A=\frac{1}{2}bh$, the integral is: $$\int_{-1}^1\,(|x|-1)\;\mathrm{d}x=\frac{1}{2}(2)(-1)=\boxed{-1}$$
$\endgroup$ $\begingroup$For $x\ge 0$, the integrand is $x-1$. The graph begins at $(0,-1)$ and ends at $(1,0)$.
For $x\le 0$, the integrand is $-x-1$. The graph begins at $(-1,0)$ and ends at $(0,-1)$.
$\endgroup$ 1 $\begingroup$The graph can be considered as a piecewise function of two lines either side of $x=0$. Specifically, these are the lines $$y=\pm x-1$$ If you graph these lines it should be clear that the integral can be found piecewise. The integral will be $$\begin{align}\int\,(|x|-1)\;\mathrm{d}x=\begin{cases}\int\,(x-1)\;\mathrm{d}x\mbox{ for }x\geq 0\\\int\,(-x-1)\;\mathrm{d}x\mbox{ for }x< 0\end{cases}\\=\begin{cases}\frac{1}{2}x^2-x+c_0\mbox{ for }x\geq 0\\-\frac{1}{2}x^2-x+c_1\mbox{ for }x< 0\end{cases}\\\end{align}$$ Where $c_0,c_1\in\mathbb{R}$
To evaluate the definite integral, then, we have $$\begin{align}\int_{-1}^1\,(|x|-1)\;\mathrm{d}x=\int_{-1}^0\,(|x|-1)\;\mathrm{d}x+\int_{0}^1\,(|x|-1)\;\mathrm{d}x\\=\left[-\frac{1}{2}x^2-x\right]_{-1}^0+\left[\frac{1}{2}x^2-x\right]_{0}^1\\=\left(\frac{1}{2}-1\right)+\left(\frac{1}{2}-1\right)\\=\boxed{-1}\end{align}$$
$\endgroup$
