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please help! :(c3 maths

JUSt iv (the using this result bit only ))-can somebody helppp? How do i approach this bit and do it as logically as possible without complicating things? the paper this is from http://mei.org.uk/files/papers/2012_Jan_c3.pdf
plu.PNG
(edited 5 years ago)
Original post by Kalabamboo
JUSt iv (the using this result bit only ))-can somebody helppp? How do i approach this bit and do it as logically as possible without complicating things? the paper this is from http://mei.org.uk/files/papers/2012_Jan_c3.pdf
plu.PNG


Well, if we reflect the shaded area in y=x, it will be an area under the graph of the inverse function, since the graph of the inverse function is the reflection of the graph of the original function in y=x. Can you see what the limits will be?
(edited 5 years ago)
Reply 2
Original post by Prasiortle
Start by performing the substitution u=2-e^x as stated.


the using this result bit only please - i have done what youve said already
Original post by Kalabamboo
the using this result bit only please - i have done what youve said already


I've edited my post (#2) now - have another look.
Reply 4
Original post by Prasiortle
Well, if we reflect the shaded area in y=x, it will be an area under the graph of the inverse function, since the graph of the inverse function is the reflection of the graph of the original function in y=x. Can you see what the limits will be?

I see what you mean here but not quite sure what limits youre referring to
Original post by Kalabamboo
I see what you mean here but not quite sure what limits youre referring to


In other words, if A is the desired area, the reflection of A in y=x will be an area B which is an area under the graph of the inverse function, i.e. an integral of the inverse function between particular limits. What are those limits?
Reply 6
Original post by Kalabamboo
JUSt iv (the using this result bit only ))-can somebody helppp? How do i approach this bit and do it as logically as possible without complicating things? the paper this is from http://mei.org.uk/files/papers/2012_Jan_c3.pdf
plu.PNG



u=2exdudx=ex12udu=dx u = 2 - e^x \Rightarrow \dfrac{du}{dx} = -e^x \Rightarrow \dfrac{-1}{2-u} du = dx
and ex=2u e^x = 2 - u
so

0log(43)g(x)dx=2e02elog(4/3)2uu12udu=12/31udu=[logu]12/3=log2/3=log3/2[br][br] \int_0 ^{\log(\dfrac{4}{3})} g(x) dx = \int_{2-e^0}^{2-e^{\log(4/3)}} \dfrac{2-u}{u} \cdot \dfrac{-1}{2-u} du = \int_{1}^{2/3} \dfrac{-1}{u} du = -[\log{u}]_{1}^{2/3} = -\log{2/3} = \log{3/2}[br] [br]
Reply 7
Original post by Ryanzmw
u=2exdudx=ex12udu=dx u = 2 - e^x \Rightarrow \dfrac{du}{dx} = -e^x \Rightarrow \dfrac{-1}{2-u} du = dx
and ex=2u e^x = 2 - u
so

0log(43)g(x)dx=2e02elog(4/3)2uu12udu=12/31udu=[logu]12/3=log2/3=log3/2[br][br] \int_0 ^{\log(\dfrac{4}{3})} g(x) dx = \int_{2-e^0}^{2-e^{\log(4/3)}} \dfrac{2-u}{u} \cdot \dfrac{-1}{2-u} du = \int_{1}^{2/3} \dfrac{-1}{u} du = -[\log{u}]_{1}^{2/3} = -\log{2/3} = \log{3/2} [br] [br]

thankBut I have already done this - what I am stuck on is the next part of iv Using this result, show that the exact area of the shaded region shown in Fig. 9 is ln 32
/27 . [Hint:]
Reply 8
Original post by Kalabamboo
thankBut I have already done this - what I am stuck on is the next part of iv Using this result, show that the exact area of the shaded region shown in Fig. 9 is ln 32
/27 . [Hint:]


Oops my bad. Draw a diagram and notice what we've computed is the area under the graph which respect to x, so the shaded area under the g to the x axis.

We can get the area of the integral of g with respect to y (which is the same as f with respect to x i.e. the area we want) by simply using the geometry of the situation:

Area of the rectangle = g(ln(4/3)) * ln(4/3) = 2*ln(4/3) = ln(16/9)
Area beneath the curve = ln(3/2)

So the area we want = ln(16/9) - ln(3/2) = ln(16/9 * (2/3)) = ln(32/27)
(edited 5 years ago)
Reply 9
Original post by Ryanzmw
Oops my bad. Draw a diagram and notice what we've computed is the area under the graph which respect to x, so the shaded area under the g to the x axis.

We can get the area of the integral of g with respect to y (which is the same as f with respect to x i.e. the area we want) by simply using the geometry of the situation:

Area of the rectangle = g(ln(4/3)) * ln(4/3) = 2*ln(4/3) = ln(16/9)
Area beneath the curve = ln(3/2)

So the area we want = ln(16/9) - ln(3/2) = ln(16/9 * (2/3)) = ln(32/27)

I am still struggling to understand this. Could you please kindly help? Finding it hard to visualise the situation and work it out.
Sure I've editted the diagram a bit:
In the integral the question makes us compute we find the shaded area (since it's with respect to x, we get the area beneath the curve to the x axis)

we can compute the area of the blue rectangle since we know the the corners

the area we want is the difference between the two (the unshaded white bit) as I detailed above! :smile:

edittedphoto.png

Original post by Kalabamboo
I am still struggling to understand this. Could you please kindly help? Finding it hard to visualise the situation and work it out.
Original post by Ryanzmw
Sure I've editted the diagram a bit:
In the integral the question makes us compute we find the shaded area (since it's with respect to x, we get the area beneath the curve to the x axis)

we can compute the area of the blue rectangle since we know the the corners

the area we want is the difference between the two (the unshaded white bit) as I detailed above! :smile:

edittedphoto.png

Thanks a lot!! :smile:

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