Question

How do you evaluate `int abs(x-1)dx` for [0, 5]?

Answer 1

`int_{0}^{5}|x-1|\ dx=17/2=8.5`

Explanation:

Note that `|x-1|=x-1` when `x geq 1` and `|x-1|=1-x` when `x<1`.

Therefore,

`int_{0}^{5} |x-1|\ dx=int_{0}^{1}(1-x)\ dx+ int_{1}^{5}(x-1)\ dx`

`=[x-x^2/2]_{0}^{1}` +

`[x^2/2-x]_{1}^{5}`

`=1-1/2+((25/2-5)-(1/2-1))`

`=1/2+15/2+1/2=17/2=8.5`.

This answer can also be found graphically by plotting the function `y=|x-1|` and adding up areas of appropriate triangles. The picture below helps us see that the integral equals `1/2 * 1 * 1+1/2 * 4 * 4=1/2+8=17/2=8.5`.

graph{|x-1| [-1.926, 8.074, -0.84, 4.16]}