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Easy

Question

The value of not stretchy integral subscript negative 2 end subscript superscript 3 end superscript vertical line 1 minus x to the power of 2 end exponent vertical line d x is‐

  1. 28 over 3    
  2. 14 over 3    
  3. 7 over 3    
  4. 1 third    

The correct answer is: 28 over 3

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not stretchy integral subscript negative pi end subscript superscript pi end superscript fraction numerator 2 x left parenthesis 1 plus blank s i n blank x right parenthesis over denominator 1 plus c o s to the power of 2 end exponent x end fraction equals

not stretchy integral subscript negative pi end subscript superscript pi end superscript fraction numerator 2 x left parenthesis 1 plus blank s i n blank x right parenthesis over denominator 1 plus c o s to the power of 2 end exponent x end fraction equals

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If I subscript n end subscript equals not stretchy integral subscript 0 end subscript superscript pi divided by 4 end superscript t a n to the power of n end exponent blank x d x then the value of n left parenthesis I subscript n minus 1 end subscript plus I subscript n plus 1 end subscript right parenthesis is‐

If I subscript n end subscript equals not stretchy integral subscript 0 end subscript superscript pi divided by 4 end superscript t a n to the power of n end exponent blank x d x then the value of n left parenthesis I subscript n minus 1 end subscript plus I subscript n plus 1 end subscript right parenthesis is‐

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Statement‐I:: not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript left parenthesis square root of x plus 1 right parenthesis d x less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis comma n element of N.
Statement‐II:: If f left parenthesis x right parenthesis is continuous and increasing in left square bracket 0 , 1 right square bracket, then not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript f left parenthesisxXlx less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis , where n element of N

Statement‐I:: not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript left parenthesis square root of x plus 1 right parenthesis d x less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction left parenthesis square root of fraction numerator r over denominator n end fraction end root plus 1 right parenthesis comma n element of N.
Statement‐II:: If f left parenthesis x right parenthesis is continuous and increasing in left square bracket 0 , 1 right square bracket, then not stretchy sum subscript r equals 0 end subscript superscript n minus 1 end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis less than not stretchy integral subscript 0 end subscript superscript 1 end superscript f left parenthesisxXlx less than not stretchy sum subscript r equals 1 end subscript superscript n end superscript fraction numerator 1 over denominator n end fraction f left parenthesis fraction numerator r over denominator n end fraction right parenthesis , where n element of N

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not stretchy integral subscript 2 end subscript superscript 3 end superscript fraction numerator square root of x over denominator square root of left parenthesis 5 minus x right parenthesis end root plus square root of x end fraction d x equals

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stack lim with n rightwards arrow infinity below invisible function application fraction numerator pi over denominator n end fraction left square bracket blank s i n blank fraction numerator pi over denominator n end fraction plus blank s i n blank fraction numerator 2 pi over denominator n end fraction plus horizontal ellipsis plus blank s i n blank fraction numerator left parenthesis n minus 1 right parenthesis pi over denominator n end fraction right square bracket is equals to:

stack lim with n rightwards arrow infinity below invisible function application fraction numerator pi over denominator n end fraction left square bracket blank s i n blank fraction numerator pi over denominator n end fraction plus blank s i n blank fraction numerator 2 pi over denominator n end fraction plus horizontal ellipsis plus blank s i n blank fraction numerator left parenthesis n minus 1 right parenthesis pi over denominator n end fraction right square bracket is equals to:

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not stretchy integral subscript e to the power of e to the power of e end exponent end exponent end subscript superscript e to the power of e to the power of e to the power of e end exponent end exponent end exponent end superscript fraction numerator d x over denominator x blank l n blank x times blank l n blank left parenthesis blank l n blank x right parenthesis times blank l n blank left parenthesis blank l n blank left parenthesis blank l n blank x right parenthesis right parenthesis end fraction equals

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If integral subscript 0 superscript pi divided by 3 end superscript fraction numerator blank c o s blank x over denominator 3 plus 4 blank s i n blank x end fraction d x equals k blank l o g blank left parenthesis fraction numerator 3 plus 2 square root of 3 over denominator 3 end fraction right parenthesis then k is‐

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