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
B
3π4π2sinxdx
C
3π4π4sinxdx
D
3ππsinxdx

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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