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?

  • $\begingroup$ 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. $\endgroup$ – TheSimpliFire Jun 11 at 7:33
  • $\begingroup$ @Cuoredicervo I just edited the post... I'll delete my comments to reduce the noise. $\endgroup$ – PierreCarre Jun 11 at 10:40
  • $\begingroup$ 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 $\int_{3/4}^1 z\,dz+\int_1^{3/2}(2-z)\,dz=\frac{19}{32}$. $\endgroup$ – StubbornAtom Jun 11 at 13:12
  • $\begingroup$ To clarify, you can find the area/probability from a picture alone (without integration). $\endgroup$ – 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.

enter image description here

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

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


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