Quite lost at figuring out what to do with this indicator function and double integral

[baobab]

Let
[MATH] f\left(x_{1}, x_{2}\right)=\exp \left(-x_{1}^{2} / 2-x_{2}^{2} / 2+x_{2} / 3+2 \log \left(\left|x_{1}\right|+1\right)\right), \quad \text { where } x_{1}, x_{2} \in \mathbb{R} [/MATH]
Solve numerically, where I_A is the indicator function:

[MATH] \int_{\mathbb{R}^{2}} I_{A} d x_{1} d x_{2}, \quad \text { with } \quad A=\left\{\left(x_{1}, x_{2}\right): f\left(x_{1}, x_{2}\right) \geq 1 / 2\right\} [/MATH]
As mentioned in the title, I am quite lost at figuring out what to do with this. I think I do understand the concept of an Indicator function ( I^A = 1 for f(x₁,x₂) >= 1/2, and 0 otherwise ) , but i don't seem to grasp how to put things together.

PS: Sorry my last math focused class was quite a while ago...

 

[Deleted member 4993]

Let
[MATH] f\left(x_{1}, x_{2}\right)=\exp \left(-x_{1}^{2} / 2-x_{2}^{2} / 2+x_{2} / 3+2 \log \left(\left|x_{1}\right|+1\right)\right), \quad \text { where } x_{1}, x_{2} \in \mathbb{R} [/MATH]
Solve numerically, where I_A is the indicator function:

[MATH] \int_{\mathbb{R}^{2}} I_{A} d x_{1} d x_{2}, \quad \text { with } \quad A=\left\{\left(x_{1}, x_{2}\right): f\left(x_{1}, x_{2}\right) \geq 1 / 2\right\} [/MATH]
As mentioned in the title, I am quite lost at figuring out what to do with this. I think I do understand the concept of an Indicator function ( I^A = 1 for f(x₁,x₂) >= 1/2, and 0 otherwise ) , but i don't seem to grasp how to put things together.

PS: Sorry my last math focused class was quite a while ago...

You were told to solve numerically.

What method of numerical integration have you been taught?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[baobab]

Yikes... i hadn't even really thought about what "solve numerically" means! Sorry.

Turns out we were allowed to use base R functions to get an approximate solution to this problem. Using an approach like Riemann sums over two dimensions.

Thank you!