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Integrals

Short Answer Type

91.

Evaluate the following integral using substitution.
$integral subscript 0 superscript 1 fraction numerator straight x over denominator straight x squared plus 1 end fraction dx$

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92.

Evaluate the following integral using substitution.

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93.

Evaluate the following definite integral:
$integral subscript 0 superscript 1 straight x space straight e to the power of straight x squared end exponent space dx$

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94.

Evaluate the following definite integral:
$integral subscript 1 superscript 2 fraction numerator 1 over denominator straight x left parenthesis 1 plus log space straight x right parenthesis squared end fraction dx$

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95.

Evaluate the following integral using substitution
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator 1 plus cos squared straight x end fraction dx$

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96.

Evaluate the following integral:
$integral subscript straight pi over 6 end subscript superscript straight pi over 2 end superscript fraction numerator cosec space straight x space cotx over denominator 1 minus cosec squared straight x end fraction dx$

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97.

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

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98.

Evaluate the following integral:
$integral subscript 0 superscript 1 fraction numerator left parenthesis tan to the power of negative 1 end exponent straight x right parenthesis squared over denominator 1 plus straight x squared end fraction dx$

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99.

Evaluate the following integral using substitution.
$integral subscript 0 superscript 2 straight x space square root of straight x plus 2 end root dx.$

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100.
Evaluate $integral subscript negative 1 end subscript superscript 1 5 straight x to the power of 4 square root of straight x to the power of 5 plus 1 end root dx.$

Let I = $integral subscript negative 1 end subscript superscript 1 left parenthesis straight x to the power of 5 plus 1 right parenthesis to the power of 1 half end exponent left parenthesis 5 straight x to the power of 4 right parenthesis space dx$
Put x⁵ + 1 = y, ∴ 5 x⁴ dx = dy When x = –1 , y = – 1 + 1 = 0 When x = 1, y = 1 + 1 = 2
$therefore$ $straight I space equals space integral subscript 0 superscript 2 straight y to the power of 1 half end exponent dy space equals space open square brackets fraction numerator straight y to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 0 superscript 2 space equals space 2 over 3 open square brackets straight y to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 2 space equals space 2 over 3 open parentheses 2 to the power of 3 over 2 end exponent minus 0 close parentheses space equals space 2 over 3 open parentheses 2 square root of 2 close parentheses space equals space fraction numerator 4 square root of 2 over denominator 3 end fraction$