Evaluate  from Mathematics Integrals

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsShort Answer Type

221.

Show that integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis space dx space equals space 2 space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx comma space space if space straight f and g are defined as f(x) = f(a - x) and g(x) + g(a-x) = 4

146 Views

 Multiple Choice QuestionsLong Answer Type

222.

Evaluate: 
integral subscript 0 superscript 1 cot to the power of negative 1 end exponent left parenthesis 1 minus straight x plus straight x squared right parenthesis dx.

142 Views

 Multiple Choice QuestionsMultiple Choice Questions

223.

The value of integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript left parenthesis straight x cubed plus straight x space cosx space plus space tan to the power of 5 straight x plus 1 right parenthesis space dx space is

  • 0

  • 2

  • straight pi
  • straight pi
125 Views

224.

The value of integral subscript 0 superscript straight pi over 2 end superscript space log space space open parentheses fraction numerator 4 plus 3 space sinx over denominator 4 plus 3 space cosx end fraction close parentheses dx is

  • 2

  • 3 over 4
  • 0

  • 0

139 Views

Advertisement
225.

If straight f left parenthesis straight a plus straight b minus straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis comma space space then space integral subscript straight a superscript straight b straight x space straight f left parenthesis straight x right parenthesis space dx is equal to

  • fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight b minus straight x right parenthesis space dx
  • fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight b plus straight x right parenthesis space dx
  • fraction numerator straight b minus straight a over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx
  • fraction numerator straight b minus straight a over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx
108 Views

226.

The value of integral subscript 0 superscript 1 space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x minus 1 over denominator 1 plus straight x minus straight x squared end fraction close parentheses dx is 

  • 1

  • 0

  • -1

  • -1

137 Views

 Multiple Choice QuestionsLong Answer Type

Advertisement

227.

Evaluate integral subscript 0 superscript straight pi over 2 end superscript log space sinx space dx.


Let I = integral subscript 0 superscript straight pi over 2 end superscript space log space sinx space dx                             ...(1)
therefore space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript log space sin open parentheses straight pi over 2 minus straight x close parentheses space dx               open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets
or     straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript space log space cosx space dx                         ...(2)
Adding (1) and (2), we get,
      2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript left parenthesis log space sinx space plus space log space cosx right parenthesis space dx space equals space integral subscript 0 superscript straight pi over 2 end superscript log space left parenthesis sinx space cosx right parenthesis space dx

            equals space integral subscript 0 superscript straight pi over 2 end superscript log space open parentheses fraction numerator 2 space sinx space cos space straight x over denominator 2 end fraction close parentheses dx space equals space integral subscript 0 superscript straight pi over 2 end superscript log space open parentheses fraction numerator sin space 2 straight x over denominator 2 end fraction close parentheses dx

therefore space space space 2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript space log space sin space 2 straight x space dx space space minus space integral subscript 0 superscript straight pi over 2 end superscript space log space 2 space dx
therefore space space space 2 space straight I space equals space straight I subscript 1 minus space straight I subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis

where   straight I subscript 2 space equals space integral subscript 0 superscript straight pi over 2 end superscript log space 2 space dx space equals space log space 2 space integral subscript 0 superscript straight pi over 2 end superscript 1. space dx
                 equals space log space 2. space open square brackets straight x close square brackets subscript 0 superscript straight pi over 2 end superscript space equals space space log space 2. space open square brackets straight pi over 2 minus 0 close square brackets space equals space straight pi over 2 space log space 2
and  straight I subscript 1 space equals space integral subscript 0 superscript straight pi over 2 end superscript space log space sin space 2 straight x space dx
Put      2 x = t          or  straight x space equals space straight t over 2 space space space space space space space space space space space space space space space space space rightwards double arrow space space space space dx space equals space 1 half dt
When   x = 0,  t  = 0
When    straight x space equals space straight pi over 2 comma space space space straight t space equals space straight pi
therefore space space space straight I subscript 1 space equals space integral subscript 0 superscript straight pi log space sint. space 1 half dt space equals space 1 half integral subscript 0 superscript straight pi log space sint space dt space equals space 1 half integral subscript 0 superscript straight pi log space sinx space dx
                                                                               open square brackets because space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight t right parenthesis space dt close square brackets
           equals space 1 half. space 2 space integral subscript 0 superscript straight pi over 2 end superscript log. space sinx space dx space equals space integral subscript 0 superscript straight pi over 2 end superscript log space sinx space dx
                                          open square brackets because space space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 2 integral subscript 0 superscript straight a over 2 end superscript straight f left parenthesis straight x right parenthesis space dx space when space straight f left parenthesis straight a minus straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis close square brackets
therefore space space space space straight I subscript 1 space equals space 1
therefore space space space space from space left parenthesis 3 right parenthesis comma space space 2 space space straight I space equals space straight I space minus space straight pi over 2 log space 2 space space space rightwards double arrow space space straight I space equals space minus straight pi over 2 space log space 2 space
rightwards double arrow space space space integral subscript 0 superscript straight pi over 2 end superscript space log space sinx space dx space equals space minus straight pi over 2 log space 2

Cor. integral subscript 0 superscript straight pi over 2 end superscript log space cosx space dx space equals space integral subscript 0 superscript straight pi over 2 end superscript log space cos open parentheses straight pi over 2 minus straight x close parentheses dx space equals space integral subscript 0 superscript straight pi over 2 end superscript log space sinx space dx space equals space minus straight pi over 2 log space 2

111 Views

Advertisement

 Multiple Choice QuestionsShort Answer Type

228.

Evaluate the definite integral:
integral subscript 2 superscript 3 fraction numerator 1 over denominator straight x squared minus 1 end fraction dx

124 Views

Advertisement
229.

Evaluate the definite integral:
integral subscript 0 superscript 1 divided by square root of 2 end superscript fraction numerator sin to the power of negative 1 end exponent straight x over denominator left parenthesis 1 minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction dx

181 Views

 Multiple Choice QuestionsLong Answer Type

230.

Evaluate:
integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator straight a space cosx plus straight b space sinx end fraction. space straight a comma space straight b space greater than space 0

195 Views

Advertisement