LynFran one year ago Help...

1. LynFran

2. 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.. $\frac{13}{2}\quad \left(\mathrm{Decimal:\quad }\:6.5\right)$

3. DecentNabeel

any confusion @LynFran

4. LynFran

ok i know the answer is 13/2...but i have to use a method from geometry ...? im confuse...

5. LynFran

@Hero

6. LynFran

@campbell_st

7. LynFran

8. LynFran

@UsukiDoll

9. UsukiDoll

I don't know the geometry method.. I would've done the same thing as @DecentNabeel

|2x-3| is an absolute value function which when you graph looks like a V you would graph it then us the triangle formula

1/2 base times height

12. UsukiDoll

yes absolute value graphs have a V shape. ^ :)

would have to break into two triangles

14. LynFran

so how do i graph it...do i need to input the lower and upper limits to get the function..im still confuse..

|dw:1436662192706:dw|

|dw:1436662409249:dw|

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

sorry that my drawing is not really scaled ^_^"

19. LynFran

where did u get those points from? |dw:1436669704611:dw|

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

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)

22. LynFran

so i dont need to input the upper and lower limits in |2x-3|...thats what im not clear no...

naaa you just need to be able to visualize it |dw:1436662834072:dw|here are the bounds

here are the triangles you need to find the geometric areas of |dw:1436662904906:dw|

25. LynFran

ok thanks

need me to go further or you got it?

27. LynFran

i got it thanks

back to sleep then have fun zzzzzzzzzzzzzzzzzzzz

29. 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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