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Home > Application Of Integrals

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Application Of Integrals

Question
CBSEENMA12036191
Wired Faculty App

The solution for x of the equation integral subscript square root of 2 end subscript superscript straight x fraction numerator dt over denominator straight t square root of straight t squared minus 1 end root end fraction space equals straight pi over 2 space is

  • 2

  • π

  • square root of 3 divided by 2
  • 2 square root of 2

Solution
Multi-choise Question

B.

π

integral subscript square root of 2 end subscript superscript straight x space fraction numerator dt over denominator straight t square root of straight t squared minus 1 end root end fraction space equals space straight pi over 2
left square bracket sec to the power of negative 1 end exponent right square bracket subscript square root of 2 end subscript superscript straight x space equals space straight pi over 2
sec to the power of negative 1 end exponent space straight x space minus space straight pi over 4 space equals space straight pi over 2
sec to the power of negative 1 end exponent space straight x space space equals space fraction numerator 3 straight pi over denominator 4 end fraction
straight x space equals space minus space square root of 2

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Some More Questions From Application of Integrals Chapter

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ` 100 and that on a bracelet is ` 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.

Solve the following L.P.P. graphically :
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0
and x, y ≥ 0

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