Maths-
General
Easy

Question

The value of the definite integral stretchy integral subscript 0 end subscript superscript 1 end superscript   open parentheses 1 plus e to the power of negative x to the power of 2 end exponent end exponent close parentheses d x is

  1. 1    
  2. 2    
  3. 1 plus e to the power of 1 end exponent    
  4. noneofthese    

hintHint:

We are aware that differentiation is the process of discovering a function's derivative and integration is the process of discovering a function's antiderivative. Thus, both processes are the antithesis of one another. Therefore, we can say that differentiation is the process of differentiation and integration is the reverse. The anti-differentiation is another name for the integration.
Here we have given:  stretchy integral subscript 0 end subscript superscript 1 end superscript   open parentheses 1 plus e to the power of negative x to the power of 2 end exponent end exponent close parentheses d x and we have to integrate it. We will use minimum and maximum method to solve this.

The correct answer is: noneofthese


    Now we have given the function as stretchy integral subscript 0 superscript 1   open parentheses 1 plus e to the power of negative x squared end exponent close parentheses d x. Here the lower limit is 0 and upper limit is 1. We know that there are some integrals where we can use the minimum and maximum method which makes problems to solve easily. We will use one of them.
    We have:
    stretchy integral subscript 0 superscript 1   open parentheses 1 plus e to the power of negative x squared end exponent close parentheses d x
N o w space c o n s i d e r space t h e space f u n n c t i o n space f left parenthesis x right parenthesis comma space i f space i t space i s space c o n t i n u o u s space f u n c t i o n space i n space a comma b space t h e n colon
m left parenthesis b minus a right parenthesis less or equal than stretchy integral subscript a superscript b f left parenthesis x right parenthesis d x less or equal than M left parenthesis b minus a right parenthesis
H e r e colon
M equals space m a x i m i m space v a l u e space o f space f left parenthesis x right parenthesis space i n space left square bracket a comma b right square bracket
m equals space m i n i m u m space v a l u e space o f space f left parenthesis x right parenthesis space i n space left square bracket a comma b right square bracket
N o w space t h e space f u n c t i o n space w e space h a v e space i s colon
f left parenthesis x right parenthesis equals open parentheses 1 plus e to the power of negative x squared end exponent close parentheses space
I t space i s space c o n t i n u o u s space i n space left square bracket 0 comma 1 right square bracket.
T h e space l i m i t s space a r e space f r o m space 0 space t o space 1 comma space s o colon
i f space 0 space less than space x space less than space 1 comma space t h e n
x squared less than x
e to the power of x squared end exponent less than e to the power of x
e to the power of negative x squared end exponent less than e to the power of negative x end exponent
i f space 0 space less than space x space less than space 1 comma space t h e n
x squared greater than 0
e to the power of x squared end exponent greater than e to the power of 0
e to the power of negative x squared end exponent less than 1
S o space w e space g e t colon
e to the power of negative x end exponent less than e to the power of negative x squared end exponent less than 1 space f o r space a l l space x element of left square bracket 0 comma 1 right square bracket.
N o w space a d d i n g space 1 space t o space a l l space t h r e e space t e r m s comma space w e space g e t colon
1 plus e to the power of negative x end exponent less than 1 plus e to the power of negative x squared end exponent less than 1 plus 1
A d d i n g space i n t e g r a l space s i g n comma space w e space g e t colon
stretchy integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x end exponent right parenthesis d x less than stretchy integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x squared end exponent right parenthesis d x less than stretchy integral subscript 0 superscript 1 2 d x
A f t e r space p u t t i n g space t h e space l i m i t s comma space w e space g e t colon
2 minus 1 over e less than stretchy integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x squared end exponent right parenthesis d x less than 2

    So here we used the concept of integrals and simplified it. We can also solve it manually but it will take lot of time to come to final answer hence we used the minimum and maximum method which makes problem to solve easily. So the answer is non of these.

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