The given curve crosses the x-axis whenever x is a multiple of π, and it lies above the x-axis between consecutive even and odd multiples of π. So the regions with area S₀, S₁, S₂, ... are the sets
![R_0 = \left\{(x, y) : 0 \le x \le \pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}](https://tex.z-dn.net/?f=R_0%20%3D%20%5Cleft%5C%7B%28x%2C%20y%29%20%3A%200%20%5Cle%20x%20%5Cle%20%5Cpi%20%5Ctext%7B%20and%20%7D%200%20%5Cle%20y%20%5Cle%20e%5E%7B-x%7D%5Csin%28x%29%5Cright%5C%7D)
![R_1 = \left\{(x, y) : 2\pi \le x \le 3\pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}](https://tex.z-dn.net/?f=R_1%20%3D%20%5Cleft%5C%7B%28x%2C%20y%29%20%3A%202%5Cpi%20%5Cle%20x%20%5Cle%203%5Cpi%20%5Ctext%7B%20and%20%7D%200%20%5Cle%20y%20%5Cle%20e%5E%7B-x%7D%5Csin%28x%29%5Cright%5C%7D)
![R_2 = \left\{(x, y) : 4\pi \le x \le 5\pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}](https://tex.z-dn.net/?f=R_2%20%3D%20%5Cleft%5C%7B%28x%2C%20y%29%20%3A%204%5Cpi%20%5Cle%20x%20%5Cle%205%5Cpi%20%5Ctext%7B%20and%20%7D%200%20%5Cle%20y%20%5Cle%20e%5E%7B-x%7D%5Csin%28x%29%5Cright%5C%7D)
and so on, with
![R_k = \left\{(x, y) : 2k\pi \le x \le (2k+1)\pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}](https://tex.z-dn.net/?f=R_k%20%3D%20%5Cleft%5C%7B%28x%2C%20y%29%20%3A%202k%5Cpi%20%5Cle%20x%20%5Cle%20%282k%2B1%29%5Cpi%20%5Ctext%7B%20and%20%7D%200%20%5Cle%20y%20%5Cle%20e%5E%7B-x%7D%5Csin%28x%29%5Cright%5C%7D)
for natural number k.
The areas themselves are then given by the integral
![S_k = \displaystyle \int_{2k\pi}^{(2k+1)\pi} \int_0^{e^{-x}\sin(x)} dy \, dx = \int_{2k\pi}^{(2k+1)\pi} e^{-x}\sin(x) \, dx](https://tex.z-dn.net/?f=S_k%20%3D%20%5Cdisplaystyle%20%5Cint_%7B2k%5Cpi%7D%5E%7B%282k%2B1%29%5Cpi%7D%20%5Cint_0%5E%7Be%5E%7B-x%7D%5Csin%28x%29%7D%20dy%20%5C%2C%20dx%20%3D%20%5Cint_%7B2k%5Cpi%7D%5E%7B%282k%2B1%29%5Cpi%7D%20e%5E%7B-x%7D%5Csin%28x%29%20%5C%2C%20dx)
Integrate by parts twice. Take
![u = e^{-x} \implies du = -e^{-x} \, dx](https://tex.z-dn.net/?f=u%20%3D%20e%5E%7B-x%7D%20%5Cimplies%20du%20%3D%20-e%5E%7B-x%7D%20%5C%2C%20dx)
![dv = \sin(x) \, dx \implies v = -\cos(x)](https://tex.z-dn.net/?f=dv%20%3D%20%5Csin%28x%29%20%5C%2C%20dx%20%5Cimplies%20v%20%3D%20-%5Ccos%28x%29)
so that
![\displaystyle \int e^{-x}\sin(x) \, dx = -e^{-x}\cos(x) - \int e^{-x}\cos(x) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20e%5E%7B-x%7D%5Csin%28x%29%20%5C%2C%20dx%20%3D%20-e%5E%7B-x%7D%5Ccos%28x%29%20-%20%5Cint%20e%5E%7B-x%7D%5Ccos%28x%29%20%5C%2C%20dx)
then
![u = e^{-x} \implies du = -e^{-x} \, dx](https://tex.z-dn.net/?f=u%20%3D%20e%5E%7B-x%7D%20%5Cimplies%20du%20%3D%20-e%5E%7B-x%7D%20%5C%2C%20dx)
![dv = \cos(x) \, dx \implies v = \sin(x)](https://tex.z-dn.net/?f=dv%20%3D%20%5Ccos%28x%29%20%5C%2C%20dx%20%5Cimplies%20v%20%3D%20%5Csin%28x%29)
so that
![\displaystyle \int e^{-x}\cos(x) \, dx = e^{-x}\sin(x) + \int e^{-x}\sin(x) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20e%5E%7B-x%7D%5Ccos%28x%29%20%5C%2C%20dx%20%3D%20e%5E%7B-x%7D%5Csin%28x%29%20%2B%20%5Cint%20e%5E%7B-x%7D%5Csin%28x%29%20%5C%2C%20dx)
Overall, we find
![\displaystyle \int e^{-x}\sin(x) \, dx = -e^{-x}\cos(x) - e^{-x}\sin(x) - \int e^{-x}\sin(x) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20e%5E%7B-x%7D%5Csin%28x%29%20%5C%2C%20dx%20%3D%20-e%5E%7B-x%7D%5Ccos%28x%29%20-%20e%5E%7B-x%7D%5Csin%28x%29%20-%20%5Cint%20e%5E%7B-x%7D%5Csin%28x%29%20%5C%2C%20dx)
or
![\displaystyle \int e^{-x}\sin(x) \, dx = -\frac12 e^{-x} (\cos(x)+\sin(x)) + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Cint%20e%5E%7B-x%7D%5Csin%28x%29%20%5C%2C%20dx%20%3D%20-%5Cfrac12%20e%5E%7B-x%7D%20%28%5Ccos%28x%29%2B%5Csin%28x%29%29%20%2B%20C)
Using the antiderivative and the fundamental theorem of calculus, we compute the k-th area to be
![\displaystyle S_k = -\frac12 e^{-(2k+1)\pi} (\cos((2k+1)\pi)+\sin((2k+1)\pi)) + \frac12 e^{-2k\pi} (\cos(2k\pi)+\sin(2k\pi))](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S_k%20%3D%20-%5Cfrac12%20e%5E%7B-%282k%2B1%29%5Cpi%7D%20%28%5Ccos%28%282k%2B1%29%5Cpi%29%2B%5Csin%28%282k%2B1%29%5Cpi%29%29%20%2B%20%5Cfrac12%20e%5E%7B-2k%5Cpi%7D%20%28%5Ccos%282k%5Cpi%29%2B%5Csin%282k%5Cpi%29%29)
![\displaystyle S_k = \frac12 e^{-2k\pi} \cos(2k\pi) \left(e^{-\pi} + 1\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S_k%20%3D%20%5Cfrac12%20e%5E%7B-2k%5Cpi%7D%20%5Ccos%282k%5Cpi%29%20%5Cleft%28e%5E%7B-%5Cpi%7D%20%2B%201%5Cright%29)
![\displaystyle S_k = \frac{e^{-\pi}+1}2 e^{-2k\pi}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S_k%20%3D%20%5Cfrac%7Be%5E%7B-%5Cpi%7D%2B1%7D2%20e%5E%7B-2k%5Cpi%7D)
Since
, the sum we want is a convergent geometric sum. As n goes to ∞, we have
![\displaystyle \lim_{n\to\infty} \sum_{k=0}^n S_k = \frac{e^{-\pi}+1}2 \cdot \frac1{1 - e^{-2\pi}} = \boxed{\frac{e^\pi+1}{4\sinh(\pi)}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Csum_%7Bk%3D0%7D%5En%20S_k%20%3D%20%5Cfrac%7Be%5E%7B-%5Cpi%7D%2B1%7D2%20%5Ccdot%20%5Cfrac1%7B1%20-%20e%5E%7B-2%5Cpi%7D%7D%20%3D%20%5Cboxed%7B%5Cfrac%7Be%5E%5Cpi%2B1%7D%7B4%5Csinh%28%5Cpi%29%7D%7D)
Answer:
<u>1.16 m (approximately)</u>
Step-by-step explanation:
Let x be the width of the path.
Total area = (5+2x)(3+2x) = 39
Expand and simplify
4x²+ 16x+15-39 = 0
4x² + 16x - 24 = 0
Simplify
x² + 4x -6 = 0
Rational factoring does not work, so use quadratic formula
x = +/- sqrt(10) -2
= 1.16 or -5.16 (reject)
= <u>1.16 m (approximately)</u>
-2 -10 -3
-7 0 -8
-6 -5 -4
That's the square, it was a fun puzzle :)
Answer: D:(-1,4)
R:(-4,4)
Step-by-step explanation:
I hope that correct
Answer:
the cost of renting the pavilion for 8 hours = $130
Step-by-step explanation:
$50 plus $10 per hour
so 8 hour = 10 * 8 = $80
$50 + $80 = $130