anonymous
  • anonymous
Evaluate the integral by interpreting it in terms of areas. S^0 on the bottom -6 (5+root36-x^2)dx I have no idea how to work this problem out, please help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ganeshie8
  • ganeshie8
Take a good look at the integrand, does it look familiar to you ?
anonymous
  • anonymous
the -6 is on the bottom of the integral, um no not really
ganeshie8
  • ganeshie8
like this ?\[\large \int\limits_{-6}^0 \color{blue}{5+\sqrt{36-x^2}}\, dx\]

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anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
May be rearrange the integrand a bit \[\large y=\color{blue}{5+\sqrt{36-x^2}}\] \[\large (y\color{blue}{-5})^2=(\color{blue}{\sqrt{36-x^2}})^2\] \[\large \color{blue}{x^2}+(y\color{blue}{-5})^2=\color{blue}{36}\] what about now ? seen this before ?
anonymous
  • anonymous
it has to do with circles right?
ganeshie8
  • ganeshie8
|dw:1436074459700:dw|
ganeshie8
  • ganeshie8
look at only the top half, because the original function is \(y=5+\sqrt{36-x^2}\), which represents only the upper half of circle.
anonymous
  • anonymous
okay
ganeshie8
  • ganeshie8
you need to find the area under that curve between x=-6 and x=0
ganeshie8
  • ganeshie8
|dw:1436074742051:dw|
ganeshie8
  • ganeshie8
thats the area the given integral represents see if you can work it
anonymous
  • anonymous
would my answer be 36pi/2? Im not really sure what Im supposed to do, this is the firts problem i have that is like this
ganeshie8
  • ganeshie8
how did you get 36pi/2 ?
anonymous
  • anonymous
1/2pir^2=1/2pi6^2=36pi/2 and then i did it another way and got 30
ganeshie8
  • ganeshie8
nope, notice that the area is made up of a "quarter circle" and a "rectangle" : |dw:1436075697415:dw|
ganeshie8
  • ganeshie8
find area of quarter circle, find area of rectangle add them up
anonymous
  • anonymous
okay, what do i do (5+root36-x^2)
anonymous
  • anonymous
would my answer be 6+9pi? @ganeshie8
ganeshie8
  • ganeshie8
whats the area of rectangle ?
ganeshie8
  • ganeshie8
|dw:1436076828225:dw|
anonymous
  • anonymous
is that a 5? so it would be 25
ganeshie8
  • ganeshie8
careful, you need to use area of rectangle formula not area of square
anonymous
  • anonymous
w*l=6*6=36
ganeshie8
  • ganeshie8
try again
anonymous
  • anonymous
30
ganeshie8
  • ganeshie8
Yes, save that, next find the area of quarter circle
anonymous
  • anonymous
1/4PI*6^2=9pi
anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
Looks good! add them up
anonymous
  • anonymous
30+9pi
ganeshie8
  • ganeshie8
\[\large \int\limits_{-6}^0 \color{blue}{5+\sqrt{36-x^2}}\, dx~~=6\times 5+\dfrac{\pi\times 6^2}{4} = 30+9\pi\]
anonymous
  • anonymous
Thank you!!!

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