#area between 3 curves (calc2)

33 messages · Page 1 of 1 (latest)

leaden rain
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hey guys!! i've attempted this problem like 8 thousand times and watched so many videos but i just cant get it. can someone help?

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fervent adderBOT
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wraith wigeon
leaden rain
# wraith wigeon well what have you tried?

so i did 4/x=2x and 4/x=x/2 to find the intersection points, which are sqrt2 and sqrt8 (respectively), then integrated from 0 to sqrt2 of (4/x-2x) to get 0.6137.. then integrated (2x-x/2) from sqrt2 to sqrt8 which was 3.5, and i added 3.5 + 0.6137 to get 4.11

wraith wigeon
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where'd 2/x come from??

leaden rain
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my bad i meant x/2 !!

wraith wigeon
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Ok

wraith wigeon
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not one of the lines and the hyperbola

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as evident from drawing the region

leaden rain
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ohh ok

wraith wigeon
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yeah

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draw regions

leaden rain
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so should i integrate from [0,sqrt8] and [sqrt2,sqrt8] instead?

wraith wigeon
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what

leaden rain
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sorry 😭

wraith wigeon
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it's clear that, on x in [0,sqrt2], that the bands are between y=2x and y=x/2

leaden rain
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OH thats what u meant my bad

wraith wigeon
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so... ask for clarification...

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but anyway, it's clear on [sqrt2,sqrt8] that the bands go from x/2 to 4/x

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hence $A=\int_0^{\sqrt{2}}2x-0.5x\dd{x}+\int_{\sqrt{2}}^{\sqrt{8}}\frac{4}{x}-.5x\dd{x}$

hexed widgetBOT
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Omegabet_

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leaden rain
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ahh ok! lemme try that rq

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i got 1.7726 but it says its still wrong

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for the first integral i got 1.5, and the second i got 0.2775887

wraith wigeon
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use exact values

leaden rain
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still not workin

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wraith wigeon
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yeah looks wrong

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$\int_{\sqrt{2}}^{\sqrt{8}}\frac{4}{x}-0.5x\dd{x}=[4\ln(x)-\frac{1}{4}x^2]_{\sqrt{2}}^{\sqrt{8}}=4\ln(\sqrt{8})-2-4\ln(\sqrt{2})+\frac{1}{2}$

hexed widgetBOT
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Omegabet_

186109587366215691.png
leaden rain
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THANK YOU!!