Binomial Theorem

Question
CBSEENMA11015614

The value of integral subscript 1 superscript straight a left square bracket straight x right square bracket straight f apostrophe left parenthesis straight x right parenthesis space dx comma space straight a space greater than 1,where [x] denotes the greatest integer not exceeding x is

  • af(a) − {f(1) + f(2) + … + f([a])}

  • [a] f(a) − {f(1) + f(2) + … + f([a])}

  • [a] f([a]) − {f(1) + f(2) + … + f(a)}

  • af([a]) − {f(1) + f(2) + … + f(a)}

Solution

B.

[a] f(a) − {f(1) + f(2) + … + f([a])}

Let a = k + h, where [a] = k and 0 ≤ h < 1
integral subscript 1 superscript straight a left square bracket straight x right square bracket straight f apostrophe left parenthesis straight x right parenthesis space dx space integral subscript 1 superscript 2 1 straight f apostrophe left parenthesis straight x right parenthesis space dx space plus integral subscript 2 superscript 3 2 straight f apostrophe space left parenthesis straight x right parenthesis dx space plus
space....... integral subscript straight k minus 1 end subscript superscript straight k left parenthesis straight k minus 1 right parenthesis space dx space plus integral subscript straight k superscript straight k plus straight h end superscript kf apostrophe left parenthesis straight x right parenthesis space dx
{f(2) − f(1)} + 2{f(3) − f(2)} + 3{f(4) − f(3)}+…….+ (k−1) – {f(k) − f(k − 1)} + k{f(k + h) − f(k)}
= − f(1) − f(2) − f(3)……. − f(k) + k f(k + h)
= [a] f(a) − {f(1) + f(2) + f(3) + …. + f([a])}

Sponsor Area

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