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04-06-2009: 4th June 2009 18:30
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#1
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Peer Of The TSR Realm
Thread Starter
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Join Date: May 2008
Location: Devon
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C4 Differentiation/Integration question
Hey guys,
Was revising earlier when I came across this question;
The equation of a curve is 
Find the co-ordinates of the turning point on the curve and determine whether it is a maximum or minimum point.
Sketch the curve.
Prove that the area of the closed region bounded by the curve, the x- and y-axes and the line  is  .
So by differentiating and solving I got the co-ordinates of the turning point to be  which is right according to the book. It's also a maximum, so I sketched the graph with some points of intersection on.
Then the integration. That's what's stumped me. I integrated by parts and got the answer 
I honestly can't see where the +2 came from. Am I wrong, or is it the book? |
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Maths revision, coursework or discussion you will find help in here.
Useful Resources:
![\int^{\ln4}_0 xe^{\frac{-x}{2}}\ dx=-2xe^{\frac{-x}{2}}-\int^{\ln4}_0 e^{\frac{-x}{2}}\ dx
=\left[-2xe^{\frac{-x}{2}}+2e^{\frac{-x}{2}}\right]_0^\ln4](http://thestudentroom.co.uk/latexrender/pictures/5ce7edc12e4ee262490b060bc99bc83f.png "\int^{\ln4}_0 xe^{\frac{-x}{2}}\ dx=-2xe^{\frac{-x}{2}}-\int^{\ln4}_0 e^{\frac{-x}{2}}\ dx
=\left[-2xe^{\frac{-x}{2}}+2e^{\frac{-x}{2}}\right]_0^\ln4")
![\left[-2xe^{\frac{-x}{2}}+2e^{\frac{-x}{2}}\right]_0^{\ln4} =(-\ln4+1)-(0+2)=-2\ln2+1-2=-2\ln2-1](http://thestudentroom.co.uk/latexrender/pictures/99f8172ebacb1532fefd37b437a40d48.png "\left[-2xe^{\frac{-x}{2}}+2e^{\frac{-x}{2}}\right]_0^{\ln4} =(-\ln4+1)-(0+2)=-2\ln2+1-2=-2\ln2-1")
