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નિયત સંકલન

નીચે આપેલાં નિયત સંકલિતને સરવાળાના લક્ષ સ્વરૂપે મેળવો :

$integral from 1 to 3 of space open parentheses straight x squared space plus space straight x close parentheses space dx$

નીચે આપેલાં નિયત સંકલિતને સરવાળાના લક્ષ સ્વરૂપે મેળવો :
$integral from negative 1 to 1 of space straight e to the power of straight x space dx$

નીચે આપેલાં નિયત સંકલિતને સરવાળાના લક્ષ સ્વરૂપે મેળવો :
$integral from 0 to 1 of space straight e to the power of 2 minus 3 straight x end exponent space dx$

નીચે આપેલાં નિયત સંકલિતને સરવાળાના લક્ષ સ્વરૂપે મેળવો :
$integral from 1 to 2 of space 3 to the power of straight x space dx$

નીચે આપેલાં નિયત સંકલિતને સરવાળાના લક્ષ સ્વરૂપે મેળવો :
$integral from log subscript straight e space 2 to log subscript straight e space 5 of space straight e to the power of straight x space dx$

અહી, વિધેય f(x) = e^xએ અંતરાલ [ log_e 2, log_e 5] માં સતત છે.

આ અંતરાલને સમાન લંબાઈ h ના n ઉપાંતરાલોમાં વિભાજન કરતાં,

પ્રત્યેક ઉપાંતરાલની લંબાઈ $straight h space equals space fraction numerator straight b space minus space straight a over denominator straight n end fraction space equals space fraction numerator log subscript straight e space 5 space minus space log subscript straight e space 2 over denominator straight n end fraction space equals space 1 over straight n space times space log subscript straight e space 5 over 2$ થશે.

તેથી $nh space equals space log subscript straight e space 5 over 2$ મળે. તથા જેમ $straight n space rightwards arrow space infinity$ તેમ $straight h space rightwards arrow space 0$ થશે.

હવે, વ્યાખ્યા પ્રમાણે,

$integral from log subscript straight e space 2 to log subscript straight e space 5 of space straight e to the power of straight x space dx space equals space limit as straight h rightwards arrow 0 of space straight h space times space sum from straight i space equals space 1 to straight n of space straight f space open parentheses log subscript straight e space 2 space plus space ih close parentheses space space space space open parentheses because space straight a space equals space log subscript straight e space 2 close parentheses$

$equals space limit as straight h rightwards arrow 0 of space straight h space sum from straight i space equals space 1 to straight n of space open parentheses straight e to the power of log subscript straight e space 2 space plus space ih end exponent close parentheses equals space limit as straight h rightwards arrow 0 of space straight h space sum from straight i space equals space 1 to straight n of space open parentheses straight e to the power of log subscript straight e space 2 end exponent space times space straight e to the power of ih close parentheses equals space limit as straight h rightwards arrow 0 of space straight h space sum from straight i space equals space 1 to straight n of space open parentheses 2 space times space straight e to the power of ih close parentheses space space open parentheses because space straight a to the power of log subscript straight a space straight y end exponent space equals space straight y close parentheses equals space limit as straight h rightwards arrow 0 of space straight h space sum from straight i space equals space 1 to straight n of space open square brackets straight e to the power of straight h space plus space straight e to the power of 2 straight h end exponent space plus space straight e to the power of 3 straight h end exponent space plus space...... plus straight e to the power of nh close square brackets equals space limit as straight h rightwards arrow 0 of space straight h space sum from straight i space equals space 1 to straight n of space open square brackets 2 straight h space times fraction numerator straight e to the power of straight h space open parentheses straight e to the power of nh space minus space 1 close parentheses over denominator straight e to the power of straight h space minus space 1 end fraction close square brackets space space space space open parentheses because space straight S subscript straight n space equals space fraction numerator straight a open parentheses straight r to the power of straight n space minus space 1 close parentheses over denominator straight r space minus space 1 end fraction space semicolon space straight r space not equal to space 1 close parentheses equals space limit as straight h rightwards arrow 0 of space straight h space sum from straight i space equals space 1 to straight n of space open square brackets fraction numerator 1 over denominator begin display style fraction numerator straight e to the power of straight h space minus space 1 over denominator straight h end fraction end style end fraction space times space 2 space straight e to the power of straight h space open parentheses straight e to the power of log subscript straight e space end subscript space end exponent 5 over 2 space minus space 1 close parentheses close square brackets$

$equals space open square brackets 1 over 1 space times space 2 space times space straight e to the power of 0 space open parentheses 5 over 2 space minus space 1 close parentheses close square brackets space space space space space space open parentheses because space straight a to the power of log space straight a to the power of straight y space equals space straight y close parentheses equals space 1 open parentheses 2 close parentheses space open parentheses 1 close parentheses space open parentheses 3 over 2 close parentheses equals space bold 3$

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