3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 :

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.