Question

The value of $\underset{n\to \infty }{lim}{\Sigma }_1^n\sin (\frac{\pi }{4}+\frac{\pi i}{2n})\frac{\pi }{2n}=?$

• A. ${\int }_{\frac{\pi }{2}}^{\frac{\pi }{4}}\sin xdx$
• B. ${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{4}}\sin xdx$
• C. ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\sin xdx$
• D. ${\int }_{\pi }^{3\pi }\sin xdx$

Answer 1

The correct option is C ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\sin xdx$

Given : $\underset{n\to \infty }{lim}{\sum }_1^n\cos (\frac{\pi }{2}+\frac{\pi i}{2n})\frac{\pi }{2n}=?$

$\underset{n\to \infty }{lim}{\sum }_1^n\sin (\frac{\pi }{4}+\frac{\pi i}{2n})\frac{\pi }{2n}=?\Delta x=\left(\frac{\pi }{2n}\right)=\frac{b-a}{n}a=\frac{\pi }{4}b=\frac{3\pi }{4}f\left(x\right)=\sin x{x}^{\ast }a+i\Delta x=\frac{\pi }{4}+\frac{\pi i}{2n}f\left(x^{\ast }\right)=\sin (\frac{\pi }{4}+\frac{\pi i}{2n})\underset{n\to \infty }{lim}{\sum }_1^n\sin (\frac{\pi }{4}+\frac{\pi i}{2n})\frac{\pi }{2n}={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\sin xdx$

Hence the correct answer is ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\sin xdx$