Yes it's exactly due to symmetry. Specifically, symmetry about the x axis.
Simply writing ![\displaystyle \int_{-8}^{-4}\left[\sqrt{x+8} \ \right]dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-8%7D%5E%7B-4%7D%5Cleft%5B%5Csqrt%7Bx%2B8%7D%20%5C%20%5Cright%5Ddx)
, without the 2 out front, will only get you the area of the portion shown in red (see diagram below).
The blue region has equal area of the red region due to symmetry.
So,
![\displaystyle 2\int_{-8}^{-4}\left[\sqrt{x+8} \ \right]dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%5Cint_%7B-8%7D%5E%7B-4%7D%5Cleft%5B%5Csqrt%7Bx%2B8%7D%20%5C%20%5Cright%5Ddx)
represents the red and blue regions combined.
The portion in green is ![\displaystyle \int_{-4}^{8}\left[\sqrt{x+8} - \frac{x}{2} \ \right]dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-4%7D%5E%7B8%7D%5Cleft%5B%5Csqrt%7Bx%2B8%7D%20-%20%5Cfrac%7Bx%7D%7B2%7D%20%5C%20%5Cright%5Ddx)
which is the integral of the difference of the upper and lower curves over the interval 
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In all honesty, it's probably easier to integrate with respect to y since the given functions are in terms of y initially. Also, there isn't a junction point in which the curves swap places in terms of which one is larger. However, it doesn't hurt to have practice in integrating with respect to x.
If you're curious about what the y integral looks like, then it would be
![\displaystyle \int_{-2}^{4} \bigg[ (2y) - (y^2-8)\bigg] dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-2%7D%5E%7B4%7D%20%5Cbigg%5B%20%282y%29%20-%20%28y%5E2-8%29%5Cbigg%5D%20dy)
You can use a tool like WolframAlpha to check that both integral expressions result in 36 to help confirm that they represent the same overall area (just in different ways of course).
Answer:
$11.10
Step-by-step explanation:
1 small package = $2.40
1 large package = $2.90
1 small package + 3 large packages.
2.40 + (3 × 2.90)
=2.40 + 8.70
=11.1
Answer:
10
Step-by-step explanation
The earthquake measures 6.4 on the Richter scale which struck Japan in Jullu 2007 and caused and extensive damage. Earlier that year, a minor earthquake measuring 3.1 in the Richter scale has stroked in parts of Pennsylvania.
Fomular:
The magnitude of an earthquake is M log(I/S)
where I donates the intensity of the earthquake and S be the intensity of the standard earthquake.
Calculation:
Consider that M1 be the magnitude Japanese earthquake and M2 be the magnitude of the Pennsylvania earthquake and L1 be the intensity of the Japanese earthquake and L2 the intensity of the Pennsylvania earthquake.
Here the magnitude of the Japanese earthquake is M1 = 6.14 and the magnitude of the Pennsylvania is M2 = 3.1
By the use of magnitude of the earthquake fomular M = log I1/S, the intensity of the Japanese earthquake is calculated as follows .
M1 = log I1/S
I1/s = 10
Answer:
6,8,
Step-by-step explanation:
