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Integrals

Question
CBSEENMA12032373

Evaluate integral subscript straight a superscript straight b cosx space dx as the limit of a sum.

Solution

Comparing integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space with space integral subscript straight a superscript straight b cosx space dx comma we get,
                 straight f left parenthesis straight x right parenthesis space equals space cos space straight x
Now,            integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space Lt with straight h rightwards arrow 0 below space straight h space left square bracket straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus... plus straight f left parenthesis straight a plus stack straight n minus 1 with bar on top right parenthesis right square bracket
 therefore            integral subscript straight a superscript straight b cosx space dx space equals space Lt with straight h rightwards arrow 0 below space straight h left square bracket cos space straight a plus space cos space left parenthesis straight a plus straight h right parenthesis plus space cos left parenthesis straight a plus 2 straight h right parenthesis plus... plus cos left parenthesis straight a plus stack straight n minus 1 with bar on top space straight h right parenthesis right square bracket
                                       equals space Lt with straight h rightwards arrow 0 below space straight h open square brackets fraction numerator cos open parentheses straight a plus begin display style fraction numerator nh minus straight h over denominator 2 end fraction end style close parentheses space sin begin display style hn over 2 end style over denominator sin begin display style straight h over 2 end style end fraction close square brackets
                                     equals space Lt with straight h rightwards arrow 0 below space open parentheses fraction numerator begin display style straight h over 2 end style over denominator sin begin display style straight h over 2 end style end fraction close parentheses space open square brackets 2 space cos space open parentheses straight a plus fraction numerator straight b minus ah over denominator 2 end fraction close parentheses space sin fraction numerator straight b minus straight a over denominator 2 end fraction close square brackets
equals space left parenthesis 1 right parenthesis space open square brackets 2 space cos open parentheses straight a plus fraction numerator straight b minus straight a minus 0 over denominator 2 end fraction close parentheses sin fraction numerator straight b minus straight a over denominator 2 end fraction close square brackets
equals space 2 space cos fraction numerator straight b plus straight a over denominator 2 end fraction sin fraction numerator straight b minus straight a over denominator 2 end fraction
equals space sinb space minus space sina

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