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Question

integral subscript negative 1 end subscript superscript 1   x e to the power of x d x equals

  1. e
  2. 1/e
  3. straight e squared
  4. 2 divided by straight e

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:  integral subscript negative 1 end subscript superscript 1   x e to the power of x d xand we have to integrate it. We will use the formula to find the answer.

The correct answer is: 2 divided by straight e


    Now we have given the function as integral subscript negative 1 end subscript superscript 1   x e to the power of x d x. Here the lower limit is -1 and upper limit is 1. We know that there are some integrals of  special functions which makes problem to solve easily. We will use one of them.
    We will use the formula:
    integral e to the power of x d x equals e to the power of x plus C
fraction numerator d left parenthesis x right parenthesis over denominator d x end fraction space equals space 1
    Now we will use the following formulas, and we get:
    integral x e to the power of x d x equals x integral e to the power of x d x minus integral left square bracket fraction numerator d x over denominator d x end fraction cross times integral e to the power of x d x right square bracket d x
x e to the power of x minus integral left parenthesis 1 cross times e to the power of x right parenthesis space d x space
x e to the power of x minus integral e to the power of x d x
x e to the power of x minus e to the power of x plus C
e to the power of x space left parenthesis x minus 1 right parenthesis space plus space C
A p p l y i n g space t h e space l i m i t s comma space w e space g e t colon
e to the power of 1 space left parenthesis 1 minus 1 right parenthesis space minus e to the power of left parenthesis negative 1 right parenthesis end exponent space left parenthesis negative 1 minus 1 right parenthesis
0 minus 1 over e left parenthesis negative 2 right parenthesis
integral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals 2 over e

    So here we used the concept of integrals of special functions 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 formula to solve. The integral of the given function isintegral subscript negative 1 end subscript superscript 1 x e to the power of x d x equals 2 over e

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    integral subscript 1 superscript 2   fraction numerator d x over denominator square root of 1 plus x squared end root end fraction equals

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    integral subscript pi divided by 4 end subscript superscript pi divided by 2 end superscript   cot invisible function application x d x equals

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