Thread: Unusual differentiation View Single Post
 2004-08-19, 16:40 #9 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 1000000001002 Posts 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