Evaluate: from Mathematics Integrals

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 Multiple Choice QuestionsShort Answer Type

231. space F i n d space colon integral fraction numerator left parenthesis 2 x minus 5 right parenthesis e to the power of 2 x end exponent over denominator left parenthesis 2 x minus 3 right parenthesis end fraction d x
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232. Find space colon integral fraction numerator straight x squared plus straight x plus 1 over denominator left parenthesis straight x squared plus 1 right parenthesis left parenthesis straight x plus 2 right parenthesis end fraction dx
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233. Evaluate space colon space integral subscript negative 2 end subscript superscript 2 fraction numerator straight x squared over denominator 1 plus 5 to the power of straight x end fraction dx.
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234. Find colon space integral left parenthesis straight x plus 3 right parenthesis square root of 3 minus 4 straight x minus straight x squared space end root space dx.
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235.

Find the integrating factor for the following differential equation:straight x space log space straight x dy over dx space plus space straight y space equals space 2 logx space

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236.

Evaluate: integral subscript 0 superscript straight pi divided by 2 end superscript fraction numerator sin squared straight x over denominator sinx plus cosx end fraction dx.

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237.

Evaluate integral subscript negative 1 end subscript superscript 2 left parenthesis straight e to the power of 3 straight x end exponent plus 7 straight x minus 5 right parenthesis dx

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238.

Evaluate:
integral fraction numerator straight x squared over denominator straight x to the power of 4 plus straight x squared minus 2 end fraction dx


integral fraction numerator straight x squared over denominator straight x to the power of 4 plus straight x squared minus 2 end fraction dx
equals integral fraction numerator straight x squared over denominator left parenthesis straight x squared minus 1 right parenthesis thin space left parenthesis straight x squared plus 2 right parenthesis end fraction dx
equals integral fraction numerator straight x squared over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 2 right parenthesis end fraction dx
Using space partial space fraction comma
space space fraction numerator straight x squared over denominator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x squared plus 2 right parenthesis end fraction space equals space fraction numerator straight A over denominator left parenthesis straight x minus 1 right parenthesis end fraction plus fraction numerator straight B over denominator left parenthesis straight x plus 1 right parenthesis end fraction plus fraction numerator Cx plus straight D over denominator left parenthesis straight x squared plus 2 right parenthesis end fraction
fraction numerator straight x squared over denominator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis left parenthesis straight x squared plus 2 right parenthesis end fraction equals fraction numerator straight A left parenthesis straight x plus 1 right parenthesis left parenthesis straight x squared plus 2 right parenthesis plus straight B left parenthesis straight x squared plus 2 right parenthesis left parenthesis straight x minus 1 right parenthesis plus left parenthesis Cx plus straight D right parenthesis left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis over denominator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis left parenthesis straight x squared plus 2 right parenthesis end fraction

Equating the coefficients from both the numerators we get,
 A+B+C = 0 ....(1)
  A-B+D = 1 ....(2)
  2A+2B-C = 0 ...(3)
  2A-2B-D = 0 .....(4)
Solving the above equations we get,
 straight A equals 1 over 6 comma space space straight B space equals space minus 1 over 6 comma space straight C equals space space 0 comma space straight D space equals space 2 over 3
Our space Intergral space becomes comma
integral fraction numerator straight x squared over denominator left parenthesis straight x minus 1 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis space left parenthesis straight x squared plus 2 right parenthesis end fraction dx space equals space integral fraction numerator 1 over denominator 6 left parenthesis straight x minus 1 right parenthesis end fraction minus fraction numerator 1 over denominator 6 left parenthesis straight x plus 1 right parenthesis end fraction plus fraction numerator 2 over denominator 3 left parenthesis straight x squared plus 2 right parenthesis end fraction dx
equals space 1 over 6 log left parenthesis straight x minus 1 right parenthesis minus 1 over 6 log left parenthesis straight x plus 1 right parenthesis plus 2 over 3 cross times fraction numerator 1 over denominator square root of 2 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of 2 end fraction close parentheses plus straight C
equals 1 over 6 open square brackets log left parenthesis straight x minus 1 right parenthesis minus log left parenthesis straight x plus 1 right parenthesis plus 2 square root of 2 tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of 2 end fraction close parentheses close square brackets plus straight C
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239.

Evaluate: integral fraction numerator left parenthesis straight x plus 3 right parenthesis straight e to the power of straight x over denominator left parenthesis straight x plus 5 right parenthesis cubed end fraction dx

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240.

Three schools X, Y, and Z organized a fete (mela) for collecting funds for flood victims in which they sold hand-held fans, mats and toys made from recycled material, the sale price of each being Rs. 25, Rs. 100 andRs. 50 respectively. The following table shows the number of articles of each type sold:
School/Article                     School X                   School Y                School Z
Hand-held fans                         30                         40                         35
      Mats                                  12                         15                         20
       Toys                                 70                         55                         75
Using matrices, find the funds collected by each school by selling the above articles and the total funds collected. Also write any one value generated by the above situation. 

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