Evaluate the following integral:
$integral subscript 1 superscript 2 1 over straight x squared straight e to the power of negative 1 over straight x end exponent dx$

Solution

Let I = $integral subscript 1 superscript 2 1 over straight x squared straight e to the power of 1 over straight x end exponent dx$
Put $1 over straight x space equals straight y comma space space space space space space therefore space space minus 1 over straight x squared dx space equals space dy space space space space space space space space space space rightwards double arrow space space space 1 over straight x squared dx space space equals negative dy$
When x = 1, y = 1
When x =2, y = $1 half$
$therefore space space space straight I space space equals negative integral subscript 1 superscript 1 half end superscript space straight e to the power of straight y space dx equals space minus open square brackets straight e to the power of straight y close square brackets subscript 1 superscript 1 half end superscript space equals space minus open square brackets straight e to the power of 1 half end exponent minus straight e to the power of 1 close square brackets space equals space straight e minus square root of straight e$

For Daily Free Study Material Join wiredfaculty

Integrals

Evaluate the following integral:
$integral subscript 1 superscript 2 1 over straight x squared straight e to the power of negative 1 over straight x end exponent dx$

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction