How do you find the integral of $abs(x) dx$ on the interval [-2, 1]?

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Answer 1 · 2015-04-12T20:27:32

$int_(x=-2)^1 abs(x) dx$

The easiest way to do this is to think about what this function looks like:
enter image source here

The integral is equal to the area shaded in blue
which is the sum of the areas of the two triangles

$(2xx2)/2 + (1xx1)/2$

$= 2 1/2$

Answer 2 · 2015-04-12T20:28:01

$absx = x$ for $x >=0$ and $absx = -x$ for $x<=0$ so to integrate

$int_-2^1 absx dx$ we need to split the integral into two integrals:

$int_-2^1 absx dx = int_-2^0 -x dx + int_0^1 x dx$ .

Tlhis method is useful and important to know for more complicated integrals, like: $int_0^8 abs (3x^2-17x+10)dx$ ,

But for this particular problem there is a straightforward geometric method.

Tlhe graph of $abs x$ :

graph{abs x [-3.07, 1.796, -0.313, 2.12]}

From $-2$ to $0$ we have a right triangle with base = height = 2, so its area is $1/2 2*2 = 2$

On the right, from 0 to 1, is a triangle of area $1/2$ , The integrhl is the area below the graph and above the axis, so

$int_-2^1 absx dx = 2+1/2=5/2$

(These are the values of the two integrals above.)

How do you find the integral of abs(x) dxabs(x) dx on the interval [-2, 1]?