integral over domain $1_{(x+y≥100)}$

Question

The problem is:

Compute: $\frac {1}{40^2}\int_{40}^{80}\int_{40}^{80} (100-x)1_{(x+y≥100)}dxdy$

my attempt:

$$=\frac {1}{40^2}\int_{40}^{80}\int_{100-x}^{80} (100-x) dydx = \frac {1}{40^2}\int_{40}^{80}(100-x)(x-20)dx = 36.67$$ But, the solution claims: $$\frac {1}{40^2}\int_{40}^{80}\int_{40}^{80} (100-x)1_{(x+y≥100)}dxdy = 33.33$$

Where am I going wrong? Is there a piece I am cutting off?

If you could also show me all the steps, that'd be great.

Answer 1

Hope this help. You should divide it into two parts.

The picture on the left is the desired region. The picture on the right shows the region you are integrating on.