View Single Post
Old 2004-08-19, 16:40   #9
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

1000000001002 Posts
Cool Unusual differentiation


Thank you for your query.
You are making the wrong assumption so naturally you get a conflicting answer!
PLease re-read my post. It is better to use leibniz's notation. Its less confusing.
The diff. coeff. is 1 and not x
Thus dy/dx =1
integrating y=x+c. put c=0 as the eqn is derived to be y=x
Therefore y=x
Now if we integrate once again we get integral ydx=(x^2)/2+c and put c=0
therefore Integral ydx =(x^2)/2. The limits we set at 0 to 1 now draw the line y=x
The integral denotes the area of a triangle formed by points (0,0) (1,0) and (1,1). The area of this triangle is 1/2(base * height) =(1*1)/2=1/2
Compare with the integral above viz. (x^2)/2. Here x =1, so area =1/2.
Therefore both results from (Calculus and co-ord geom) tally Q.E.D
:confused
I am open for further questions

Mally
mfgoode is offline   Reply With Quote