LAST DOUBT B4 JEE................

How to differentaite or integrate [x] ????? plzzzz tell fast i forgot [2][2][2]

11 Answers

11
Subash ·

for integration and differenciation consider separate intervals

like 0-1,1-2...

1
Akand ·

oh tht i kno.......w8 ill post d question wher i was stuck..........

1
Akand ·

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

1
The Race begins... ·

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!)

1357
Manish Shankar ·

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])

1
Akand ·

ah manish bhaiyya nice......................

awesum method.....

we luv u.....

62
Lokesh Verma ·

for derivative look the the shortcuts in the magazine ;)

basically it will be zero between integers and not exist at the integer points..

62
Lokesh Verma ·

race you can do that.. but then how do you simplifiy that expression?

1
The Race begins... ·

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. :(

1
Kalyan Pilla ·

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.

[339]

3
iitimcomin ·

and differentiating it is easy ...............

for all x belonging to integers its not valid ..............

if it dosnt belong to intgers d[x]/dx = 0,,,,

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