# Area of the region $\{(x,y):0\leq x \leq 1, 0 \leq y\leq 1, 3/4\leq x+y\leq 3/2\}$

Find the area of the region $$\{(x,y):0\leq x \leq 1, 0 \leq y\leq 1, 3/4\leq x+y\leq 3/2\}$$ (using definite integration).

I cannot understand how to find this area. I have graphed the lines and found out the required region. I found the definite integral $$\int_{0}^{1} (3/2-y)-(3/4-y)dy$$ but it is yielding an extra areas. How do I find the area of the region?

• HINT: Find the area of the unit square (easy). Then find the area enclosed by the axes and $x+y=\frac34$. Lastly find the area enclosed by $x=1$, $y=1$ and $x+y=\frac32$ and perform the appropriate subtraction. In this case it may be easier to do it without integration as the areas are simple geometric shapes. – TheSimpliFire Jun 11 at 7:33
• @Cuoredicervo I just edited the post... I'll delete my comments to reduce the noise. – PierreCarre Jun 11 at 10:40
• This is the same as the probability $P(3/4<X+Y<3/2)$ where $X,Y$ are independent uniform variables on $(0,1)$. So from this post, the area is $\int_{3/4}^1 z\,dz+\int_1^{3/2}(2-z)\,dz=\frac{19}{32}$. – StubbornAtom Jun 11 at 13:12
• To clarify, you can find the area/probability from a picture alone (without integration). – StubbornAtom Jun 11 at 13:47

Since $$y \leq 1$$, it is :

$$\frac{3}{4} \leq x + y \leq \frac{3}{2} \Rightarrow -\frac{1}{4} \leq x \leq \frac{1}{2}$$

But $$x \geq 0$$, thus :

$$0 \leq x \leq \frac{1}{2}$$

Then, for $$y$$ one would get :

$$\frac{1}{4} \leq y \leq 1$$

The desired area of $$D = \{(x,y) \in \mathbb R^2 : x \geq 1, y \leq 1, \frac{3}{4} \leq x+y \leq \frac{3}{2}\}$$, is :

$$A(D) = \iint_D \mathrm{d}x\mathrm{d}y = \int_0^\frac{1}{2} \int_{\frac{1}{4}}^1\mathrm{d}x\mathrm{d}y$$

We should be able to find the area of this polygon without calculus.

At the very least, you should be able to divide it up into a bunch of triangles.

by the shoelace algorithm I get:

$$\begin {array}{} \frac 34 & 0\\ 1&0\\ 1&\frac 12\\ \frac 12& 1\\ 0&1\\ 0&\frac 34 \end{array}$$

$$\frac 12( \frac 12 + 1 + \frac 12 - \frac 14 - \frac 9{16}) = \frac {19}{32}$$

If you want to use calculus, most direct would be

$$\int_0^{\frac 12} 1 - (\frac 34 - x) \ dx + \int_{\frac 12}^{\frac 34} (3/2 - x) - (\frac 34 - x) \ dx + \int_{\frac 34}^1 3/2 - x \ dx$$