oh tht i kno.......w8 ill post d question wher i was stuck..........
How to differentaite or integrate [x] ????? plzzzz tell fast i forgot [2][2][2]
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for integration and differenciation consider separate intervals
like 0-1,1-2...
1∫a [x] f'(x) dx
integrate this.....ive tried using by parts......but cannot get anything....
this is a ques from d forum itself.....
plzz answer this
b∫a[x]dx = (a-b)1∫2[x]dx = (a-b) ........ {a>b>0} ..... [check it's validity by substituting arbitrary values for a&b]
can we conclude in this manner? (nishant sir, please help!)
let n≤a<n+1
1∫a [x] f'(x) dx=1∫2 1.f'(x) dx+2∫3 2.f'(x) dx.....n∫a n.f'(x) dx
f(2)-f(1)+2{f(3)-f(2)}.....n{f(a)-f(n)
-f(1)-f(2)-f(3)-....-f(n)+nf(a)
now n=[a]
so it becomes
[a]f(a)-{f(1)+f(2)+....+f([a])
ah manish bhaiyya nice......................
awesum method.....
we luv u.....
for derivative look the the shortcuts in the magazine ;)
basically it will be zero between integers and not exist at the integer points..
race you can do that.. but then how do you simplifiy that expression?
oh yes, i'm sorry. actually i imagined the graph of {x} in my mind while solving. [x] will not have same area under curve b/w two integers rite.!! sorry sorry. :(
From the figure
Area under the curve is........
1+2+3+4........[a]+[a]{a} = [a]/2 ([a]+1) +[a]{a}
Thats the easiest way to integrate this, I guess.
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and differentiating it is easy ...............
for all x belonging to integers its not valid ..............
if it dosnt belong to intgers d[x]/dx = 0,,,,