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Question

The value of $$\displaystyle \lim_{n \rightarrow \infty} \Sigma_1^n \sin\left(\dfrac{\pi}{4} +\dfrac{\pi i}{2n} \right) \dfrac{\pi}{2n}=?$$


A
π4π2sinxdx
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B
3π4π2sinxdx
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C
3π4π4sinxdx
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D
3ππsinxdx
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Solution

The correct option is C $$\displaystyle \int_{\tfrac{\pi}{4}}^{\tfrac{3\pi}{4}} \sin x dx$$
Given : $$\underset { n\rightarrow \infty  }{ lim } \sum _{ 1 }^{ n }{ \cos { \left( \cfrac { \pi  }{ 2 } +\cfrac { \pi i }{ 2n }  \right) \cfrac { \pi  }{ 2n }  }  } =?$$
$$\underset { n\rightarrow \infty  }{ lim } \sum _{ 1 }^{ n }{ \sin { \left( \cfrac { \pi  }{ 4 } +\cfrac { \pi i }{ 2n }  \right) \cfrac { \pi  }{ 2n }  }  } =?\\ \Delta x=\left( \cfrac { \pi  }{ 2n }  \right) =\cfrac { b-a }{ n } \\ a=\cfrac { \pi  }{ 4 } \quad b=\cfrac { 3\pi  }{ 4 } \\ f(x)=\sin { x } \\ { x }^{ * }a+i\Delta x=\cfrac { \pi  }{ 4 } +\cfrac { \pi i }{ 2n } \\ f({ x }^{ * })=\sin { \left( \cfrac { \pi  }{ 4 } +\cfrac { \pi i }{ 2n }  \right)  } \\ \underset { n\rightarrow \infty  }{ lim } \sum _{ 1 }^{ n }{ \sin { \left( \cfrac { \pi  }{ 4 } +\cfrac { \pi i }{ 2n }  \right) \cfrac { \pi  }{ 2n }  }  } =\int _{ \cfrac { \pi  }{ 4 }  }^{ \cfrac { 3\pi  }{ 4 }  }{ \sin { x } dx } $$
Hence the correct answer is $$\int _{ \cfrac { \pi  }{ 4 }  }^{ \cfrac { 3\pi  }{ 4 }  }{ \sin { x } dx } $$

Mathematics

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