sketch the graph of the integrand function and use it to help evaluate the integral. 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 
 A: Another method would be to use the graph alone. 
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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:
$$\int_{-1}^1\,(|x|-1)\;\mathrm{d}x=\frac{1}{2}(2)(-1)=\boxed{-1}$$
A: 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 $(0,-1)$.
A: 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
$$\begin{align}\int_{-1}^1\,(|x|-1)\;\mathrm{d}x=\int_{-1}^0\,(|x|-1)\;\mathrm{d}x+\int_{0}^1\,(|x|-1)\;\mathrm{d}x\\=\left[-\frac{1}{2}x^2-x\right]_{-1}^0+\left[\frac{1}{2}x^2-x\right]_{0}^1\\=\left(\frac{1}{2}-1\right)+\left(\frac{1}{2}-1\right)\\=\boxed{-1}\end{align}$$
