Question

Help...

Answer 1

\[\mathrm{Find\:the\:equivalent\:expressions\:\to\:}\left|2x-3\right|\mathrm{\:at\:}-1\le \:x\le \:2\mathrm{\:without\:the\:absolutes}\]
\[-1\le \:x\le \frac{3}{2}:\quad \left(3-2x\right)\]
\[\frac{3}{2}\le \:x\le \:2:\quad \left(2x-3\right)\]
\[=\int\limits _{-1}^{\frac{3}{2}}\left(3-2x\right)dx+\int\limits _{\frac{3}{2}}^2\left(2x-3\right)dx\]
\[\int\limits _{-1}^{\frac{3}{2}}\left(3-2x\right)dx=\frac{25}{4}\]
\[\int\limits _{\frac{3}{2}}^2\left(2x-3\right)dx=\frac{1}{4}\]
\[=\frac{25}{4}+\frac{1}{4}\]
\[=\frac{13}{2}\]
Answer is..
\[\frac{13}{2}\quad \left(\mathrm{Decimal:\quad }\:6.5\right)\]

Answer 2

any confusion @LynFran

Answer 3

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

Answer 4

@Hero

Answer 5

@campbell_st

Answer 6

@TheSmartOne @abb0t @sleepyhead314

Answer 7

@UsukiDoll

Answer 8

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

Answer 9

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

Answer 10

1/2 base times height

Answer 11

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

Answer 12

would have to break into two triangles

Answer 13

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