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