11for integration and differenciation consider separate intervals
like 0-1,1-2...
How to differentaite or integrate [x] ????? plzzzz tell fast i forgot [2][2][2]
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11∫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
1b∫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!)
1357let 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])
1ah manish bhaiyya nice......................
awesum method.....
we luv u.....
62for derivative look the the shortcuts in the magazine ;)
basically it will be zero between integers and not exist at the integer points..
62race you can do that.. but then how do you simplifiy that expression?
1oh 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. :(
1
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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3and differentiating it is easy ...............
for all x belonging to integers its not valid ..............
if it dosnt belong to intgers d[x]/dx = 0,,,,