Answer: Option A
:
A
We have seen that there are functions which have discontinuity but still can be integrated. These functions have finite type discontinuities. Please try to understand one thing here, that definite integral gives you the area so area can be calculated by the curve into pieces as well.
For example -

We can see that the given function is discontinuous at x =1. So to integrate this function from (-3,3) we’ll have to integrate it into pieces. One piece will be (-3,1) and another will be (1, 3).
${\int }_{-3}^{3}f\left(x\right)dx={\int }_{-3}^{1}f\left(x\right)dx+{\int }_1^{3}f\left(x\right)dx$
And given that for $(-3,1)f\left(x\right)=x^2$
And for (1,3) f(x) = 6-x
${\int }_{-3}^{3}f\left(x\right)dx={\int }_{-3}^1\left(x^2\right)dx+{\int }_1^3(6-x)dx$
This way we can calculate the area even if the function is discontinuous.