Question

How to calculate this integral `int_-2^2 f(x)dx`?

Where #f(x) = {(x+2 if x \ge 0),(-x^2+2 if x<0):}#
Now I've calculated
`int_-2^0 -x^2+2dx` + `int_0^2 x+2dx` = `22/3`
Is the answer right?

Now let's call this function `g(x) = -x^2+2` and in the definition of `f(x)` is used only for `x < 0`, why we calculate `int_-2^(0) g(x)` and not `int_-2^(0^-) g(x)`?
I mean `0` is not included in the interval

Answer 1

Please see below.

Explanation:

I don't know what your notation on the upper limit of `0^-` means.

In any event, for this function we could have just as easily said

`f(x) = {(x+2,"if", x >= 0),(-x^2+2,"if", x <= 0):}`

because either rule makes `f(0) = 2`.

Moreover

As long as there is no vertical asymptote at the endpoint of the interval, the function need not be continuous on a closed interval in order to be integrable.

All necessary limits exist and are the same for

`F(x) = {(x+2,"if", x > 0),(9,"if",x = 0),(-x^2+2,"if", x < 0):}`

as for the function you asked about.

So `int_-2^2 F(x) dx = int_-2^2 f(x) dx`

And even

`G(x) = {(x+2,"if", x > 0),(9,"if",x = 0),(-x^2+5,"if", x < 0):}`

is integrable on `[-2,2]`. Because the limits needed for integrability do exist.