Question

How do you evaluate the definite integral `int abs(x^2-4)dx` from [0,3]?

Answer 1

`int_0^3 |x^2-4|dx = 23/3`

Explanation:

This is the graph of `y=x^2-4`:
graph{x^2-4 [-10, 10, -5, 5]}
This is the graph of `y=|x^2-4|`:
graph{|x^2-4| [-10, 10, -5, 5]}

It should be clear that `|x^2-4|=0=>x^2=4=>x=+=2`

So,
`\ \ \ \ \ int_0^3 |x^2-4|dx = int_0^2-(x^2-4)dx + int_2^3(x^2-4)dx`

`:. int_0^3 |x^2-4|dx = int_2^3(x^2-4)dx - int_0^2(x^2-4)dx`

`:. int_0^3 |x^2-4|dx = [1/3x^3-4x]_2^3 - [1/3x^3-4x]_0^2`

`:. int_0^3 |x^2-4|dx = {(1/3*27-4*3)-(1/3*8-4*2)} - {(1/3*8-4*2)-0}`
`:. int_0^3 |x^2-4|dx = {(9-12)-(8/3-8)} - {(8/3-8)}`
`:. int_0^3 |x^2-4|dx = {(-3)-(-16/3)} - {(-16/3)}`
`:. int_0^3 |x^2-4|dx = -3 +16/3+16/3`
`:. int_0^3 |x^2-4|dx = 23/3`

Answer 2

Without considering the graph of the integrand, see below.

Explanation:

We are asked to evaluate `int_0^3 |x^2-4|dx`.

We know that

`|x^2-4| = {(x^2-4,if, x^2-4 >= 0),(-(x^2-4),if,x^2-4 < 0) :}`

Investigate the sign of #`(x^2-4)` `on` `[0,3]`#

`x^2-4= 0` at `x= +- 2` so we need `x = 2` (We are only interested in values of `x` between `0` and `3`)

A sign analysis reveals that `x^2-4` is negative on `[0,2]` and positive on `[2, 3]`.

Split up the integral

`int_0^3 |x^2-4|dx = int_0^2 -(x^2-4) dx +int_2^3 (x^2-4) dx`

`= (16/3) + (7/3)= 23/3`

I have a preference for the preceding method. As a student, I used a version of the following.

As soon as we learn that the zeros are `+-2` we can integrate from 0 to 2 and also integrate from 2 to 3. If we get a negative answer, we'll make it positive. Then add the positive numbers together.

Formally, we use

`int_0^3 |x^2-4|dx = abs(int_0^2 (x^2-4) dx) + abs(int_2^3 (x^2-4) dx)`

`= abs(-16/3)+abs(7/3) = 23/16`.