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Integrals

Short Answer Type

61.

Evaluate the following definite integral:
$integral subscript 0 superscript 2 fraction numerator 5 straight x plus 1 over denominator straight x squared plus 4 end fraction dx$

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62.
Evaluate the following definite integral:
$integral subscript 0 superscript 1 fraction numerator 2 straight x plus 3 over denominator 5 straight x squared plus 1 end fraction dx$

Let I = $integral subscript 0 superscript 1 fraction numerator 2 straight x plus 3 over denominator 5 straight x squared plus 1 end fraction dx$
$equals space 2 integral subscript 0 superscript 1 fraction numerator xdx over denominator 5 straight x squared plus 1 end fraction plus 3 integral subscript 0 superscript 1 fraction numerator 1 over denominator 5 straight x squared plus 1 end fraction dx space equals space 2 over 10 integral subscript 0 superscript 1 fraction numerator 10 xdx over denominator 5 straight x squared plus 1 end fraction plus 3 over 5 integral subscript 0 superscript 1 fraction numerator 1 over denominator straight x squared plus begin display style 1 fifth end style end fraction dx$
$equals space 1 fifth integral subscript 0 superscript 1 fraction numerator 10 straight x over denominator 5 straight x squared plus 1 end fraction dx plus 3 over 5 integral subscript 0 superscript 1 fraction numerator 1 over denominator straight x squared plus open parentheses begin display style fraction numerator 1 over denominator square root of 5 end fraction end style close parentheses squared end fraction dx$
$equals space 1 fifth open square brackets log left parenthesis 5 straight x squared plus 1 right parenthesis close square brackets subscript 0 superscript 1 plus 3 over 5. fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator square root of 5 end fraction end style end fraction open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator begin display style fraction numerator 1 over denominator square root of 5 end fraction end style end fraction close parentheses close square brackets subscript 0 superscript 1$
$equals space 1 fifth left parenthesis log space 6 minus log 1 right parenthesis plus fraction numerator 3 over denominator square root of 5 end fraction left square bracket tan to the power of negative 1 end exponent square root of 5 minus tan to the power of negative 1 end exponent 0 right square bracket equals space 1 fifth left parenthesis log space 6 space minus 0 right parenthesis plus fraction numerator 3 over denominator square root of 5 end fraction left square bracket tan to the power of negative 1 end exponent square root of 5 minus 0 right square bracket space equals space 1 fifth log space 6 space plus space fraction numerator 3 over denominator square root of 5 end fraction tan to the power of negative 1 end exponent square root of 5$

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63.

Evaluate the following definite integral:
$integral subscript 1 superscript 2 left parenthesis log space straight x squared right parenthesis dx.$

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64.

Evaluate the following definite integral:
$integral subscript 0 superscript straight pi over 2 end superscript space straight x squared space sinx space dx.$

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65.

Evaluate the following definite integral:
$integral subscript 0 superscript straight pi over 4 end superscript space sin space 2 straight x space sin space 3 straight x space dx.$

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66.

Evaluate the following definite integral
$integral subscript 0 superscript straight pi over 4 end superscript space sin cubed space 2 straight t space cos space 2 straight t space dt$

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67.

By using the properties of definite integrals, evaluate the following integral.
$integral subscript 0 superscript straight pi over 2 end superscript space cos squared space dx$

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68.

Prove the following:
$integral subscript 0 superscript straight pi over 2 end superscript space sin cubed straight x space dx space equals space 2 over 3$