Yes the answer is -2. The value of integral between 0 and 1 is negative, as f(x) is under the x axis.
The value of integral from 1 to 3 is positive as f(x) is above the x axis.
Whenever the integral is positive its value is equal to the area.
Whenever the integral is negative its value is equal...
dy/dt=1-t^-2
dx/dt=1+t^-2
dy/dx=(dy/dt)/(dx/dt)
so dy/dx=(1-t^-2)/(1+t^-2) Multiply the top and the bottom by t^2 to get:
or dy/dx=(t^2-1)/(t^2+1) differentiate RHS with respect to t then multiply by dt/dx (the chain rule)
The second derivative= (2t(t^2+1)-2t(t^2-1))/(t^2+1)^2x t^2/(t^2+1)...
Make the t the subject in both equations you get: t=x/2 and t=1/y then
equate the right hand sides to get: x/2=1/y cross multiply to get:
xy=2 then make y the subject to have y=2/x or y=2x^-1
Differentiate to get the 1st derivative dy/dx=-2x^-2
differentiate again to get the second derivative...
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This might be of interest to physics students:
Stars eruption seen 170 years ago from earth can be seen again after being reflected. For more details visit the link below.
http://uanews.org/node/44892
Re: Tuition mathematics,r Physics (Cambridge Coaching St george & Hurstville)
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If you do physics, this might interest you. Once there was an ocean on the surface of Mars.
http://www.astronomy.com/en/News-Observing/News/2012/02/Mars Express radar gives strong evidence for former Mars ocean.aspx
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Something which might be interesting for physics students is the fact that a German sattelite of about 2500kg returned to surface missing narrowly parts of China. Google it for more info.
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