Author Topic: Finding the area question help (Read 1091 times) Tweet Share

jaiezou · « **on:** April 04, 2018, 01:11:03 pm »

Hey guys,

I wondering how I would answer this question: find the area bounded by y=5lnx and the x=6

RuiAce · « **Reply #1 on:** April 04, 2018, 01:13:03 pm »

Have you drawn a diagram to help you do this?

jaiezou · « **Reply #2 on:** April 04, 2018, 01:28:37 pm »

Quote from: RuiAce on April 04, 2018, 01:13:03 pm

Have you drawn a diagram to help you do this?

Yea I got this from desmos (couldn’t upload due to jpeg). But I can’t do the question because we can’t integrate natural log equations right? So I’m kind of at a dead end here.

RuiAce · « **Reply #3 on:** April 04, 2018, 01:46:27 pm »

Quote from: jaiezou on April 04, 2018, 01:28:37 pm

Yea I got this from desmos (couldn’t upload due to jpeg). But I can’t do the question because we can’t integrate natural log equations right? So I’m kind of at a dead end here.

$\text{Yep, we can't integrate }\log.$
$\text{The idea is that we can't do the question if}\\ \text{we keep thinking about it with respect to the }x\text{-axis.}\\ \text{We adjust our analysis by considering areas to the right}\\ \text{of the }y\text{-axis instead.}$
$\text{Rearranging }y = 5\ln x \text{ gives }x = e^{y/5}.\\ \text{Also note that when }x=1, y=0.\\ \text{And when }x=6, y=5\ln 6.$
______________________________________
$\text{What we require is the area enclosed under }y = 5 \ln x.\\ \text{Draw a rectangle around the points }(0,0), (0,5\ln 6), (6,5\ln 6), (6, 0).\\ \text{Observe that the area we require forms one part of the rectangle.}$
$\text{On the other hand, the }\textbf{other}\text{ area is given by}\\ \int_0^{5\ln 6} e^{y/5}\,dy.\\ \text{Hence, by considering a difference of areas, we work around}\\ \text{this issue by computing }\boxed{A = 6\times 5\ln 6 - \int_0^{5\ln 6} e^{y/5}\,dy}$

jaiezou · « **Reply #4 on:** April 04, 2018, 01:50:38 pm »

Quote from: RuiAce on April 04, 2018, 01:46:27 pm

$\text{Yep, we can't integrate }\log.$
$\text{The idea is that we can't do the question if}\\ \text{we keep thinking about it with respect to the }x\text{-axis.}\\ \text{We adjust our analysis by considering areas to the right}\\ \text{of the }y\text{-axis instead.}$
$\text{Rearranging }y = 5\ln x \text{ gives }x = e^{y/5}.\\ \text{Also note that when }x=1, y=0.\\ \text{And when }x=6, y=5\ln 6.$
______________________________________
$\text{What we require is the area enclosed under }y = 5 \ln x.\\ \text{Draw a rectangle around the points }(0,0), (0,5\ln 6), (6,5\ln 6), (6, 0).\\ \text{Observe that the area we require forms one part of the rectangle.}$
$\text{On the other hand, the }\textbf{other}\text{ area is given by}\\ \int_0^{5\ln 6} e^{y/5}\,dy.\\ \text{Hence, by considering a difference of areas, we work around}\\ \text{this issue by computing }\boxed{A = 6\times 5\ln 6 - \int_0^{5\ln 6} e^{y/5}\,dy}$

Thanks for ur ingenious solution!

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