Evaluate $integral subscript negative 1 end subscript superscript 1 space straight e to the power of straight x space dx$ as the limit of a sum.

Solution

Comparing $integral subscript negative 1 end subscript superscript 1 space straight e to the power of straight x space dx. space with space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx comma space we space get comma$
$straight f left parenthesis straight x right parenthesis space equals space straight e to the power of straight x comma space space space straight a space equals space minus 1 comma space space straight b space equals space 1$
$therefore space space straight f left parenthesis straight a right parenthesis space equals space straight e to the power of straight a comma space space straight f left parenthesis straight a plus straight h right parenthesis space equals space straight e to the power of straight a plus straight h end exponent comma space space straight f left parenthesis straight a plus 2 straight h right parenthesis space equals space straight e to the power of straight a plus 2 straight h end exponent comma space space..... comma$
$straight f left parenthesis straight a plus stack straight n minus 1 with bar on top space straight h right parenthesis space equals space straight e to the power of straight a plus left parenthesis straight n minus 1 right parenthesis straight h end exponent$
Now $integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space Lt with straight h rightwards arrow 0 below space straight h left square bracket straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus... plus straight f left parenthesis straight a plus stack straight n minus 1 with bar on top space straight h right parenthesis right square bracket$
$therefore$ $integral subscript negative 1 end subscript superscript 1 space straight e to the power of straight x space dx space equals space Lt with straight h rightwards arrow 0 below space straight h space open square brackets straight e to the power of straight a plus straight e to the power of straight a plus straight h end exponent plus straight e to the power of straight a plus 2 straight h end exponent plus.... plus straight e to the power of straight a plus left parenthesis straight n minus 1 right parenthesis space straight h end exponent close square brackets$
$equals space Lt with straight h rightwards arrow 0 below straight h open square brackets fraction numerator straight e to the power of straight a left parenthesis straight e to the power of straight n space straight h end exponent minus 1 right parenthesis over denominator straight e to the power of straight h minus 1 end fraction close square brackets space equals space Lt with straight h rightwards arrow 0 below straight h open square brackets fraction numerator straight e to the power of straight a left parenthesis straight e squared minus 1 right parenthesis over denominator straight e to the power of straight h minus 1 end fraction close square brackets$
$left square bracket because space space straight n space straight h space equals space straight b space minus space straight a space equals space 1 plus 1 space equals space 2 right square bracket$
$equals straight e to the power of straight a left parenthesis straight e squared minus 1 right parenthesis space Lt with straight h rightwards arrow 0 below fraction numerator straight h over denominator straight e to the power of straight h minus 1 end fraction space equals straight e to the power of straight a left parenthesis straight e squared minus 1 right parenthesis space fraction numerator 1 over denominator begin display style Lt with straight h rightwards arrow 0 below fraction numerator straight e to the power of straight h minus 1 over denominator straight h end fraction end style end fraction$
$equals space straight e to the power of straight a left parenthesis straight e squared minus 1 right parenthesis. space 1 over 1 space equals space straight e to the power of straight a left parenthesis straight e squared minus 1 right parenthesis space equals space straight e to the power of negative 1 end exponent left parenthesis straight e squared minus 1 right parenthesis space space space space space left square bracket because space straight a space equals space minus 1 right square bracket equals space straight e minus straight e to the power of negative 1 end exponent space equals space straight e minus 1 over straight e$

For Daily Free Study Material Join wiredfaculty

Integrals

Evaluate $integral subscript negative 1 end subscript superscript 1 space straight e to the power of straight x space dx$ as the limit of a sum.

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction