LynFran
  • LynFran
Help...
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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LynFran
  • LynFran
1 Attachment
DecentNabeel
  • DecentNabeel
\[\mathrm{Find\:the\:equivalent\:expressions\:\to\:}\left|2x-3\right|\mathrm{\:at\:}-1\le \:x\le \:2\mathrm{\:without\:the\:absolutes}\] \[-1\le \:x\le \frac{3}{2}:\quad \left(3-2x\right)\] \[\frac{3}{2}\le \:x\le \:2:\quad \left(2x-3\right)\] \[=\int\limits _{-1}^{\frac{3}{2}}\left(3-2x\right)dx+\int\limits _{\frac{3}{2}}^2\left(2x-3\right)dx\] \[\int\limits _{-1}^{\frac{3}{2}}\left(3-2x\right)dx=\frac{25}{4}\] \[\int\limits _{\frac{3}{2}}^2\left(2x-3\right)dx=\frac{1}{4}\] \[=\frac{25}{4}+\frac{1}{4}\] \[=\frac{13}{2}\] Answer is.. \[\frac{13}{2}\quad \left(\mathrm{Decimal:\quad }\:6.5\right)\]
DecentNabeel
  • DecentNabeel
any confusion @LynFran

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More answers

LynFran
  • LynFran
ok i know the answer is 13/2...but i have to use a method from geometry ...? im confuse...
LynFran
  • LynFran
@Hero
LynFran
  • LynFran
@campbell_st
LynFran
  • LynFran
@TheSmartOne @abb0t @sleepyhead314
LynFran
  • LynFran
@UsukiDoll
UsukiDoll
  • UsukiDoll
I don't know the geometry method.. I would've done the same thing as @DecentNabeel
sleepyhead314
  • sleepyhead314
|2x-3| is an absolute value function which when you graph looks like a V you would graph it then us the triangle formula
sleepyhead314
  • sleepyhead314
1/2 base times height
UsukiDoll
  • UsukiDoll
yes absolute value graphs have a V shape. ^ :)
sleepyhead314
  • sleepyhead314
would have to break into two triangles
LynFran
  • LynFran
so how do i graph it...do i need to input the lower and upper limits to get the function..im still confuse..
sleepyhead314
  • sleepyhead314
|dw:1436662192706:dw|
sleepyhead314
  • sleepyhead314
|dw:1436662409249:dw|
sleepyhead314
  • sleepyhead314
we can get two triangles one with a base of (1.5 + 1) and height of 5 another with a base of (2 - 1.5) and height of 1
sleepyhead314
  • sleepyhead314
sorry that my drawing is not really scaled ^_^"
LynFran
  • LynFran
where did u get those points from? |dw:1436669704611:dw|
sleepyhead314
  • sleepyhead314
by graphing... y = |2x - 3| y intercept is "-3" coming out of absolute value +3 x intercept is 1.5 find the points given by bounds when x = -1 --> y = 5 when x = 2 --> y = 1
sleepyhead314
  • sleepyhead314
you have learned how to graph absolute value functions before... right? o-o (i mean, if you're learning calculus, I'm assuming you know how)
LynFran
  • LynFran
so i dont need to input the upper and lower limits in |2x-3|...thats what im not clear no...
sleepyhead314
  • sleepyhead314
naaa you just need to be able to visualize it |dw:1436662834072:dw|here are the bounds
sleepyhead314
  • sleepyhead314
here are the triangles you need to find the geometric areas of |dw:1436662904906:dw|
LynFran
  • LynFran
ok thanks
sleepyhead314
  • sleepyhead314
need me to go further or you got it?
LynFran
  • LynFran
i got it thanks
sleepyhead314
  • sleepyhead314
back to sleep then have fun zzzzzzzzzzzzzzzzzzzz
LynFran
  • LynFran
no wait..look at this one how they get those coordinate for that sketch ?..didnt they input the limit to get it? https://www.symbolab.com/solver/calculus-calculator/%5Cint_%7B-1%7D%5E%7B4%7Dx%20dx%20/?origin=enterkey @sleepyhead314

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